CBSE Class 12 Mathematics Linear Differential Equations Worksheet Set 01

CBSE Class 12 Mathematics Linear Differential Equations (1). The Relations And Functions questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practice them to clear their Relations And Functions concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Relations And Functions worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Relations And Functions chapter and other subjects too. Use them for better understanding of the subjects.

Some Important Results/Concepts

Order of Differential Equation: The order of the highest-order derivative present in a given differential equation is defined as the order of that differential equation.

Degree of Differential Equation: The highest power to which the highest-order derivative is raised, once the differential equation is expressed as a polynomial in its derivatives (meaning all derivative powers are positive integers), is defined as the degree of the differential equation.

Homogeneous Differential Equation: A differential equation of the first order and first degree is said to be homogeneous if it can be written in the form \(\frac{dy}{dx} = \frac{f_1(x, y)}{f_2(x, y)}\), where \(f_1(x, y)\) and \(f_2(x, y)\) are homogeneous functions of the same degree.

Linear Differential Equation:

i. A differential equation of the form \(\frac{dy}{dx} + py = q\), where \(p\) and \(q\) are functions of \(x\) or constants.
The solution of this equation is given by: \(y \cdot e^{\int p \, dx} = \int \left( e^{\int p \, dx} \cdot q \right) dx\), where \(e^{\int p \, dx}\) is the Integrating Factor (I.F.).

ii. A differential equation of the form \(\frac{dx}{dy} + px = q\), where \(p\) and \(q\) are functions of \(y\) or constants.
The solution of this equation is given by: \(x \cdot e^{\int p \, dy} = \int \left( e^{\int p \, dy} \cdot q \right) dy\), where \(e^{\int p \, dy}\) is the Integrating Factor (I.F.).

Assignments

1. Order and Degree of a Differential Equation

Level I

Question 1. Write the order and degree of the following differential equation:
(i) \( \left(\frac{d^2 y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 + 2y = 0 \)
Answer:
The given differential equation is:
\( \left(\frac{d^2 y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^3 + 2y = 0 \)
Here, the highest-order derivative present is \( \frac{d^2 y}{dx^2} \), which means the order is 2.
Since the equation is a polynomial in its derivatives, the degree is the power of this highest-order derivative, which is 2.
Therefore, Order = 2, Degree = 2.
In simple words: The highest derivative here is the second derivative, so the order is 2. The power of this second derivative is 2, so the degree is also 2.

Exam Tip: Remember that degree is only defined when the differential equation can be written as a polynomial in its derivatives. Always check for fractional powers or transcendental functions of derivatives first.

2. General and Particular Solutions of a Differential Equation

Level I

Question 1. Show that \( y = e^{-x} + ax + b \) is the solution of \( e^x \frac{d^2 y}{dx^2} = 1 \).
Answer:
We are given the function:
\( y = e^{-x} + ax + b \)
Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(e^{-x} + ax + b) = -e^{-x} + a \)
Differentiating again with respect to \( x \):
\( \frac{d^2 y}{dx^2} = \frac{d}{dx}(-e^{-x} + a) = e^{-x} \)
Now, substitute \( \frac{d^2 y}{dx^2} \) into the left-hand side of the given differential equation:
\( \text{L.H.S.} = e^x \frac{d^2 y}{dx^2} = e^x (e^{-x}) = e^{x-x} = e^0 = 1 = \text{R.H.S.} \)
Since L.H.S. = R.H.S., the given function is indeed a solution of the differential equation.
In simple words: To show this is a solution, find the second derivative of the given equation and multiply it by \( e^x \). You will get 1, which proves it is correct.

Exam Tip: When showing a relation is a solution, clearly state LHS and RHS after substitution to show they are equal.

3. Formation of Differential Equation

Level II

Question 1. Obtain the differential equation by eliminating a and b from the equation \( y = e^x (a \cos x + b \sin x) \).
Answer:
The given equation is:
\( y = e^x (a \cos x + b \sin x) \)
Dividing by \( e^x \):
\( y e^{-x} = a \cos x + b \sin x \) - (1)
Differentiating (1) with respect to \( x \) using the product rule:
\( \frac{dy}{dx} e^{-x} - y e^{-x} = -a \sin x + b \cos x \) - (2)
Differentiating (2) with respect to \( x \) again:
\( \frac{d^2 y}{dx^2} e^{-x} - \frac{dy}{dx} e^{-x} - \left( \frac{dy}{dx} e^{-x} - y e^{-x} \right) = -a \cos x - b \sin x \)
\( \left( \frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + y \right) e^{-x} = -(a \cos x + b \sin x) \)
Using equation (1), we can substitute \( -(a \cos x + b \sin x) \) with \( -y e^{-x} \):
\( \left( \frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + y \right) e^{-x} = -y e^{-x} \)
Since \( e^{-x} \neq 0 \), we can divide both sides by \( e^{-x} \):
\( \frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + y = -y \)
\( \frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \)
This is the required differential equation.
In simple words: Differentiate the given equation twice and substitute the original expression back in to get rid of the constants \(a\) and \(b\).

Exam Tip: The number of arbitrary constants (here, 2 constants: \(a\) and \(b\)) equals the order of the resulting differential equation. Hence, we must differentiate exactly twice.

Level III

Question 1. Find the differential equation of the family of circles \( (x - a)^2 + (y - b)^2 = r^2 \) where \( a \) and \( b \) are arbitrary constants.
Answer:
The equation of the family of circles is given by:
\( (x - a)^2 + (y - b)^2 = r^2 \) - (1)
Here, \( a \) and \( b \) are two arbitrary constants, and \( r \) is a fixed radius (constant).
Differentiating equation (1) with respect to \( x \):
\( 2(x - a) + 2(y - b)\frac{dy}{dx} = 0 \)
\( (x - a) + (y - b)\frac{dy}{dx} = 0 \) - (2)
From (2), we get:
\( x - a = -(y - b)\frac{dy}{dx} \) - (3)
Now, differentiating equation (2) with respect to \( x \) again:
\( 1 + \left(\frac{dy}{dx}\right)^2 + (y - b)\frac{d^2 y}{dx^2} = 0 \)
\( (y - b)\frac{d^2 y}{dx^2} = -\left[ 1 + \left(\frac{dy}{dx}\right)^2 \right] \)
\( y - b = -\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2 y}{dx^2}} \) - (4)
Substituting the value of \( y - b \) from (4) into (3):
\( x - a = -\left( -\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2 y}{dx^2}} \right) \frac{dy}{dx} = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right] \frac{dy}{dx}}{\frac{d^2 y}{dx^2}} \) - (5)
Now, substitute the values of \( x - a \) and \( y - b \) from equations (5) and (4) into the original equation (1):
\( \left( \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right] \frac{dy}{dx}}{\frac{d^2 y}{dx^2}} \right)^2 + \left( -\frac{1 + \left(\frac{dy}{dx}\right)^2}{\frac{d^2 y}{dx^2}} \right)^2 = r^2 \)
\( \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^2 \left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2 y}{dx^2}\right)^2} + \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^2}{\left(\frac{d^2 y}{dx^2}\right)^2} = r^2 \)
\( \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^2 \left[ \left(\frac{dy}{dx}\right)^2 + 1 \right]}{\left(\frac{d^2 y}{dx^2}\right)^2} = r^2 \)
\( \left[1 + \left(\frac{dy}{dx}\right)^2\right]^3 = r^2 \left(\frac{d^2 y}{dx^2}\right)^2 \)
This is the required differential equation.
In simple words: Find the first and second derivatives, express the center terms \(x - a\) and \(y - b\) using these derivatives, and put them back into the circle's equation to remove the center variables.

Exam Tip: For problems involving the family of circles of a fixed radius \( r \), the final differential equation should contain \( r \) but must be free from the arbitrary constants \( a \) and \( b \).

Question 2. Obtain the differential equation representing the family of parabola having vertex at the origin and axis along the positive direction of x-axis.
Answer:
The equation of the family of parabolas having their vertex at the origin and axis along the positive direction of the x-axis is:
\( y^2 = 4ax \) - (1)
where \( a \) is an arbitrary constant (the parameter).
Differentiating equation (1) with respect to \( x \):
\( 2y\frac{dy}{dx} = 4a \) - (2)
Substituting the value of \( 4a \) from (2) into (1):
\( y^2 = \left( 2y\frac{dy}{dx} \right) x \)
Since \( y \neq 0 \) for a non-trivial parabola, we can divide both sides by \( y \):
\( y = 2x\frac{dy}{dx} \)
\( 2x\frac{dy}{dx} - y = 0 \)
This is the required differential equation representing the family of parabolas.
In simple words: Write the basic equation of the parabola, \( y^2 = 4ax \). Differentiate it to get the value of \( 4a \), and then replace \( 4a \) in the first equation to get the final differential equation.

Exam Tip: Since there is only one arbitrary constant \( a \), the order of the resulting differential equation must be exactly 1.

4. Solution of Differential Equation by the Method of Separation of Variables

Level II

Question 1. Solve \( \frac{dy}{dx} = 1 + x + y + xy \).
Answer:
The given differential equation is:
\( \frac{dy}{dx} = 1 + x + y(1 + x) \)
Factorizing the right-hand side:
\( \frac{dy}{dx} = (1 + x)(1 + y) \)
Separating the variables:
\( \frac{1}{1 + y} dy = (1 + x) dx \)
Integrating both sides:
\( \int \frac{1}{1+y} \, dy = \int (1+x) \, dx \)
\( \log|1 + y| = x + \frac{x^2}{2} + C \)
where \( C \) is the constant of integration.
In simple words: Group the terms on the right side to factor it as \( (1+x)(1+y) \). Then move the \(y\) terms to one side and the \(x\) terms to the other side, and integrate both sides.

Exam Tip: Always look for factoring opportunities in separation of variables problems to make separating \(x\) and \(y\) easier.

Question 2. Solve \( \frac{dy}{dx} = e^{-y} \cos x \) given that \( y(0) = 0 \).
Answer:
The given differential equation is:
\( \frac{dy}{dx} = e^{-y} \cos x \)
Separating the variables by bringing all terms of \( y \) to LHS and \( x \) to RHS:
\( e^y dy = \cos x \, dx \)
Integrating both sides:
\( \int e^y \, dy = \int \cos x \, dx \)
\( e^y = \sin x + C \) - (1)
Now, apply the given initial condition \( y(0) = 0 \) (i.e., \( y = 0 \) when \( x = 0 \)):
\( e^0 = \sin 0 + C \)
\( 1 = 0 + C \implies C = 1 \)
Substitute \( C = 1 \) back into equation (1):
\( e^y = \sin x + 1 \)
Taking the natural logarithm of both sides:
\( y = \log(\sin x + 1) \)
This is the required particular solution.
In simple words: Separate the variables to get \( e^y dy = \cos x dx \) and integrate. Use the condition \( y(0) = 0 \) to find the value of the constant \( C \), which turns out to be 1.

Exam Tip: When given an initial condition, always substitute the values at the integration step to find the particular constant \( C \) rather than leaving it in the general form.

Question 3. Solve \( (1 + x^2)\frac{dy}{dx} - x = \tan^{-1} x \).
Answer:
The given differential equation is:
\( (1 + x^2)\frac{dy}{dx} = x + \tan^{-1} x \)
Dividing both sides by \( 1 + x^2 \):
\( \frac{dy}{dx} = \frac{x + \tan^{-1} x}{1 + x^2} \)
Separating the variables:
\( dy = \left( \frac{x}{1 + x^2} + \frac{\tan^{-1} x}{1 + x^2} \right) dx \)
Integrating both sides:
\( y = \int \frac{x}{1 + x^2} \, dx + \int \frac{\tan^{-1} x}{1 + x^2} \, dx \) - (1)
Let us solve the two integrals on the RHS:
For the first integral, let \( 1 + x^2 = u \implies 2x \, dx = du \implies x \, dx = \frac{du}{2} \):
\( \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \log(1 + x^2) \)
For the second integral, let \( \tan^{-1} x = t \implies \frac{1}{1 + x^2} \, dx = dt \):
\( \int \frac{\tan^{-1} x}{1 + x^2} \, dx = \int t \, dt = \frac{t^2}{2} = \frac{(\tan^{-1} x)^2}{2} \)
Substituting these results back into equation (1):
\( y = \frac{1}{2} \log(1 + x^2) + \frac{1}{2}(\tan^{-1} x)^2 + C \)
where \( C \) is the constant of integration.
In simple words: Rearrange the equation so that \( \frac{dy}{dx} \) is by itself on one side. Then separate the variables and integrate each piece using substitution.

Exam Tip: Be comfortable with integration by substitution, as it is frequently required after separating the variables.

5. Homogeneous Differential Equation of First Order and First Degree

Level II

Question 1. Solve \( (x^2 + xy) \, dy = (x^2 + y^2) \, dx \).
Answer:
The given equation can be written as:
\( \frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} \) - (1)
This is a homogeneous differential equation of first order and first degree.
Let \( y = vx \). Differentiating with respect to \( x \):
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into (1):
\( v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{x^2 + x(vx)} = \frac{x^2(1 + v^2)}{x^2(1 + v)} \)
\( v + x\frac{dv}{dx} = \frac{1 + v^2}{1 + v} \)
\( x\frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v \)
\( x\frac{dv}{dx} = \frac{1 + v^2 - v - v^2}{1 + v} = \frac{1 - v}{1 + v} \)
Separating the variables:
\( \frac{1 + v}{1 - v} \, dv = \frac{dx}{x} \)
Rewriting the LHS:
\( \frac{-(1 - v) + 2}{1 - v} \, dv = \left( -1 + \frac{2}{1 - v} \right) dv \)
Integrating both sides:
\( \int \left( -1 + \frac{2}{1 - v} \right) dv = \int \frac{1}{x} \, dx \)
\( -v - 2\log|1 - v| = \log|x| + C' \)
\( -v = \log|x| + \log|1 - v|^2 + C' \)
\( -v = \log|x(1 - v)^2| + C' \)
Substituting back \( v = \frac{y}{x} \):
\( -\frac{y}{x} = \log\left| x\left(1 - \frac{y}{x}\right)^2 \right| + C' \)
\( -\frac{y}{x} = \log\left| \frac{(x-y)^2}{x} \right| + C' \)
\( \frac{y}{x} = \log\left| \frac{x}{(x-y)^2} \right| + C \)
\( e^{y/x} = \frac{C x}{(x-y)^2} \implies (x - y)^2 = C x e^{-y/x} \)
where \( C = e^{-C'} \) is an arbitrary constant.
In simple words: Since this is homogeneous, substitute \( y = vx \). This turns the equation into a separable form, which we can integrate and then substitute \( v = y/x \) back.

Exam Tip: For homogeneous equations of the form \( \frac{dy}{dx} = f(x, y) \), always use the standard substitution \( y = vx \), which reduces it to a variable-separable equation in \( v \) and \( x \).

Level III

Question 1. Show that the given differential equation is homogeneous and solve it: \( (x - y)\frac{dy}{dx} = x + 2y \).
Answer:
The given differential equation is:
\( \frac{dy}{dx} = \frac{x + 2y}{x - y} \) - (1)
Let \( f(x, y) = \frac{x + 2y}{x - y} \). To show it is homogeneous, we replace \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda x + 2\lambda y}{\lambda x - \lambda y} = \frac{\lambda(x + 2y)}{\lambda(x - y)} = f(x, y) \)
Thus, the equation is a homogeneous differential equation of degree 0.
To solve it, let \( y = vx \). Differentiating with respect to \( x \):
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into (1):
\( v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx} = \frac{1 + 2v}{1 - v} \)
\( x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v \)
\( x\frac{dv}{dx} = \frac{1 + 2v - v(1 - v)}{1 - v} = \frac{v^2 + v + 1}{1 - v} \)
Separating variables:
\( \frac{1 - v}{v^2 + v + 1} \, dv = \frac{dx}{x} \)
Integrating both sides:
\( \int \frac{1 - v}{v^2 + v + 1} \, dv = \int \frac{1}{x} \, dx \) - (2)
We rewrite the numerator of the LHS as \( 1 - v = -\frac{1}{2}(2v + 1) + \frac{3}{2} \):
\( \int \frac{-\frac{1}{2}(2v + 1) + \frac{3}{2}}{v^2 + v + 1} \, dv = -\frac{1}{2} \int \frac{2v + 1}{v^2 + v + 1} \, dv + \frac{3}{2} \int \frac{1}{\left(v + \frac{1}{2}\right)^2 + \frac{3}{4}} \, dv \)
\( = -\frac{1}{2}\log|v^2 + v + 1| + \frac{3}{2} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) \)
\( = -\frac{1}{2}\log|v^2 + v + 1| + \sqrt{3} \tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) \)
Substituting this back into (2):
\( -\frac{1}{2}\log|v^2 + v + 1| + \sqrt{3} \tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) = \log|x| + C' \)
Multiplying by 2:
\( -\log|v^2 + v + 1| + 2\sqrt{3} \tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) = 2\log|x| + C \)
\( 2\sqrt{3} \tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) = \log|x^2(v^2 + v + 1)| + C \)
Substitute \( v = \frac{y}{x} \):
\( 2\sqrt{3} \tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) = \log\left| x^2 \left( \frac{y^2}{x^2} + \frac{y}{x} + 1 \right) \right| + C \)
\( \log|x^2 + xy + y^2| - 2\sqrt{3} \tan^{-1}\left(\frac{x + 2y}{\sqrt{3}x}\right) = C \)
This is the general solution.
In simple words: Show the equation does not change when substituting \( \lambda x \) and \( \lambda y \). Then solve by substituting \( y = vx \), separating the variables, and integrating carefully.

Exam Tip: Splitting the numerator to match the derivative of the denominator is a standard technique for quadratic denominators. Don't forget to use the formula \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \).

Question 2. Show that the given differential equation is homogeneous and solve it: \( y \, dx + x \log\left(\frac{y}{x}\right) dy - 2x \, dy = 0 \).
Answer:
The given equation is:
\( y \, dx = \left[ 2x - x \log\left(\frac{y}{x}\right) \right] dy \)
\( \frac{dy}{dx} = \frac{y}{x \left[ 2 - \log\left(\frac{y}{x}\right) \right]} \) - (1)
Let \( f(x, y) = \frac{y}{x \left[ 2 - \log\left(\frac{y}{x}\right) \right]} \). Replacing \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x \left[ 2 - \log\left(\frac{\lambda y}{\lambda x}\right) \right]} = \frac{y}{x \left[ 2 - \log\left(\frac{y}{x}\right) \right]} = f(x, y) \)
Hence, the equation is homogeneous.
Let \( y = vx \). Differentiating with respect to \( x \):
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into (1):
\( v + x\frac{dv}{dx} = \frac{vx}{x(2 - \log v)} = \frac{v}{2 - \log v} \)
\( x\frac{dv}{dx} = \frac{v}{2 - \log v} - v = v \left( \frac{1 - (2 - \log v)}{2 - \log v} \right) = \frac{v(\log v - 1)}{2 - \log v} \)
Separating variables:
\( \frac{2 - \log v}{v(\log v - 1)} \, dv = \frac{dx}{x} \)
Integrating both sides:
\( \int \frac{2 - \log v}{v(\log v - 1)} \, dv = \int \frac{1}{x} \, dx \) - (2)
In the LHS integral, substitute \( \log v = t \implies \frac{1}{v} \, dv = dt \):
\( \int \frac{2 - t}{t - 1} \, dt = \int \frac{-(t - 2)}{t - 1} \, dt = \int \frac{-(t - 1 - 1)}{t - 1} \, dt = \int \left( -1 + \frac{1}{t - 1} \right) dt \)
\( = -t + \log|t - 1| = -\log v + \log|\log v - 1| \)
Substituting back into equation (2):
\( -\log v + \log|\log v - 1| = \log|x| + C' \)
\( \log\left| \frac{\log v - 1}{v} \right| = \log|x| + C' \)
\( \frac{\log v - 1}{v} = C x \)
Substitute \( v = \frac{y}{x} \):
\( \frac{\log(y/x) - 1}{y/x} = C x \)
\( \log\left(\frac{y}{x}\right) - 1 = C y \)
This is the general solution.
In simple words: This differential equation is homogeneous because all terms have matching powers when you count logarithms of \(y/x\) as degree 0. After making the \(y = vx\) substitution, we use \(t = \log v\) to perform the integration on the \(v\) side.

Exam Tip: For expressions containing \( \log(y/x) \), \( e^{y/x} \), or trigonometric functions of \( y/x \), the substitution \( y = vx \) is almost always the required approach.

Question 3. Solve \( x \, dy - y \, dx = \sqrt{x^2 - y^2} \, dx \).
Answer:
The given differential equation can be rewritten as:
\( x \, dy = (y + \sqrt{x^2 - y^2}) \, dx \)
\( \frac{dy}{dx} = \frac{y + \sqrt{x^2 - y^2}}{x} \) - (1)
This is a homogeneous differential equation. Let \( y = vx \). Differentiating both sides with respect to \( x \):
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into (1):
\( v + x\frac{dv}{dx} = \frac{vx + \sqrt{x^2 - v^2 x^2}}{x} = \frac{x(v + \sqrt{1 - v^2})}{x} \)
\( v + x\frac{dv}{dx} = v + \sqrt{1 - v^2} \)
\( x\frac{dv}{dx} = \sqrt{1 - v^2} \)
Separating the variables:
\( \frac{1}{\sqrt{1 - v^2}} \, dv = \frac{dx}{x} \)
Integrating both sides:
\( \int \frac{1}{\sqrt{1 - v^2}} \, dv = \int \frac{1}{x} \, dx \)
\( \sin^{-1} v = \log|x| + C \)
Substitute back \( v = \frac{y}{x} \):
\( \sin^{-1}\left(\frac{y}{x}\right) = \log|x| + C \)
This can also be written as:
\( y = x \sin(\log|x| + C) \)
which is the general solution.
In simple words: Rearrange the equation to isolate \( \frac{dy}{dx} \). Since it is homogeneous, we let \( y = vx \) and solve the simplified variable-separable equation using the sine-inverse integral.

Exam Tip: Remember the standard integral \( \int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) \). It commonly appears in homogeneous equations containing square roots of quadratic terms.

Question 4. Solve \( x^2 y \, dx - (x^3 + y^3) dy = 0 \).
Answer:
The given equation can be written as:
\( (x^3 + y^3) dy = x^2 y \, dx \)
\( \frac{dy}{dx} = \frac{x^2 y}{x^3 + y^3} \) - (1)
This is a homogeneous differential equation. Let \( y = vx \). Differentiating with respect to \( x \):
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substitute \( y = vx \) and \( \frac{dy}{dx} \) into (1):
\( v + x\frac{dv}{dx} = \frac{x^2 (vx)}{x^3 + v^3 x^3} = \frac{v}{1 + v^3} \)
\( x\frac{dv}{dx} = \frac{v}{1 + v^3} - v = v \left( \frac{1 - 1 - v^3}{1 + v^3} \right) = -\frac{v^4}{1 + v^3} \)
Separating variables:
\( \frac{1 + v^3}{v^4} \, dv = -\frac{dx}{x} \)
\( \left( v^{-4} + \frac{1}{v} \right) dv = -\frac{dx}{x} \)
Integrating both sides:
\( \int \left( v^{-4} + \frac{1}{v} \right) dv = -\int \frac{1}{x} \, dx \)
\( \frac{v^{-3}}{-3} + \log|v| = -\log|x| + C \)
\( -\frac{1}{3v^3} + \log|v| + \log|x| = C \)
\( -\frac{1}{3v^3} + \log|vx| = C \)
Substituting \( vx = y \) and \( v = \frac{y}{x} \):
\( -\frac{x^3}{3y^3} + \log|y| = C \)
\( \log|y| - \frac{x^3}{3y^3} = C \)
This is the general solution.
In simple words: Write \( \frac{dy}{dx} \) by itself, use \( y = vx \), and simplify. The equation splits into two simple parts which we integrate to find the solution.

Exam Tip: Combining logarithmic terms like \( \log|v| + \log|x| = \log|vx| = \log|y| \) simplifies the final equation significantly and is highly favored by examiners.

Question 5. Solve \( x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx \).
Answer:
The given equation can be written as:
\( x \, dy = (y + \sqrt{x^2 + y^2}) \, dx \)
\( \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} \) - (1)
Let \( y = vx \). Then \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute these into (1):
\( v + x\frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2 x^2}}{x} = v + \sqrt{1 + v^2} \)
\( x\frac{dv}{dx} = \sqrt{1 + v^2} \)
Separating variables:
\( \frac{1}{\sqrt{1 + v^2}} \, dv = \frac{dx}{x} \)
Integrating both sides:
\( \int \frac{1}{\sqrt{1 + v^2}} \, dv = \int \frac{1}{x} \, dx \)
\( \log|v + \sqrt{1 + v^2}| = \log|x| + \log|C| \)
\( \log|v + \sqrt{1 + v^2}| = \log|Cx| \)
\( v + \sqrt{1 + v^2} = Cx \)
Substitute \( v = \frac{y}{x} \):
\( \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx \)
\( \frac{y + \sqrt{x^2 + y^2}}{x} = Cx \)
\( y + \sqrt{x^2 + y^2} = Cx^2 \)
This is the general solution.
In simple words: Use the substitution \( y = vx \) to turn the equation into a separable form. Then integrate both sides using the standard formula for \( \int \frac{1}{\sqrt{x^2+a^2}} dx \) to get the final solution.

Exam Tip: The formula \( \int \frac{1}{\sqrt{x^2+a^2}} \, dx = \log|x + \sqrt{x^2+a^2}| \) is a key integration formula often tested in differential equations. Keep its form in mind.

Question 6. Solve \( (y + 3x^2)\frac{dx}{dy} = x \).
Answer:
The given equation can be rewritten by taking the reciprocal on both sides:
\( \frac{dy}{dx} = \frac{y + 3x^2}{x} \)
\( \frac{dy}{dx} = \frac{y}{x} + 3x \)
\( \frac{dy}{dx} - \frac{1}{x}y = 3x \) - (1)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where:
\( P = -\frac{1}{x} \) and \( Q = 3x \)
First, find the Integrating Factor (I.F.):
\( \text{I.F.} = e^{\int P \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x} \)
The general solution is given by:
\( y \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dx + C \)
\( y \cdot \frac{1}{x} = \int \left[ 3x \cdot \frac{1}{x} \right] \, dx + C \)
\( \frac{y}{x} = \int 3 \, dx + C \)
\( \frac{y}{x} = 3x + C \)
\( y = 3x^2 + Cx \)
where \( C \) is an arbitrary constant.
In simple words: Rearrange the equation into the standard linear form \( \frac{dy}{dx} - \frac{y}{x} = 3x \). Calculate the integrating factor \( \frac{1}{x} \), multiply it through, and integrate to find \( y \).

Exam Tip: If a differential equation looks complicated, try taking the reciprocal to see if it becomes linear in \(x\) or linear in \(y\).

Question 7. Solve \( x \, dy + (y - x^3) \, dx = 0 \).
Answer:
The given differential equation is:
\( x \, dy = (x^3 - y) \, dx \)
\( \frac{dy}{dx} = \frac{x^3 - y}{x} \)
\( \frac{dy}{dx} = x^2 - \frac{y}{x} \)
\( \frac{dy}{dx} + \frac{1}{x}y = x^2 \) - (1)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where:
\( P = \frac{1}{x} \) and \( Q = x^2 \)
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\log x} = x \)
The general solution is:
\( y \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dx + C \)
\( y \cdot x = \int (x^2 \cdot x) \, dx + C \)
\( xy = \int x^3 \, dx + C \)
\( xy = \frac{x^4}{4} + C \)
\( y = \frac{x^3}{4} + \frac{C}{x} \)
where \( C \) is an arbitrary constant.
In simple words: Rearrange the equation to \( \frac{dy}{dx} + \frac{y}{x} = x^2 \). The integrating factor is \( x \). Multiplying both sides by \( x \) makes the LHS \( d(xy) \), which is then integrated easily.

Exam Tip: Recognizing the derivative of a product \( d(xy) = x\,dy + y\,dx \) can save you time on exams without always needing the full integrating factor steps.

Question 8. Solve \( x \, dy + (y + 2x^2) \, dx = 0 \).
Answer:
The given equation can be written as:
\( x \, dy = -(y + 2x^2) \, dx \)
\( \frac{dy}{dx} = -\frac{y + 2x^2}{x} \)
\( \frac{dy}{dx} + \frac{1}{x}y = -2x \) - (1)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where:
\( P = \frac{1}{x} \) and \( Q = -2x \)
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\log x} = x \)
The general solution is:
\( y \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dx + C \)
\( xy = \int (-2x \cdot x) \, dx + C \)
\( xy = -2 \int x^2 \, dx + C \)
\( xy = -\frac{2x^3}{3} + C \)
\( y = -\frac{2x^2}{3} + \frac{C}{x} \)
where \( C \) is the constant of integration.
In simple words: Put the equation into standard linear form \( \frac{dy}{dx} + \frac{y}{x} = -2x \). The integrating factor is \( x \). Multiply and integrate to find the general solution.

Exam Tip: Be careful with negative signs when shifting terms across the equality sign.

6. Linear Differential Equations

Level I

Question 1. Find the integrating factor of the differential equation \( x\frac{dy}{dx} - y = 2x^2 \).
Answer:
The given differential equation is:
\( x\frac{dy}{dx} - y = 2x^2 \)
Dividing both sides by \( x \):
\( \frac{dy}{dx} - \frac{1}{x}y = 2x \) - (1)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where:
\( P = -\frac{1}{x} \)
The Integrating Factor (I.F.) is given by:
\( \text{I.F.} = e^{\int P \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x} \)
Therefore, the Integrating Factor is \( \frac{1}{x} \).
In simple words: Divide the whole equation by \( x \) first to get the standard form. The term in front of \( y \) is \( -\frac{1}{x} \). Integrating this and raising \( e \) to its power gives \( \frac{1}{x} \).

Exam Tip: Always make sure the coefficient of \( \frac{dy}{dx} \) is exactly 1 before identifying \( P \) to calculate the Integrating Factor.

Level II

Question 1. Solve \( \frac{dy}{dx} + 2y \tan x = \sin x \).
Answer:
The given differential equation is already in the standard form \( \frac{dy}{dx} + Py = Q \), where:
\( P = 2\tan x \) and \( Q = \sin x \)
First, find the Integrating Factor (I.F.):
\( \text{I.F.} = e^{\int P \, dx} = e^{\int 2\tan x \, dx} = e^{2\log|\sec x|} = e^{\log(\sec^2 x)} = \sec^2 x \)
The general solution is:
\( y \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dx + C \)
\( y \cdot \sec^2 x = \int (\sin x \cdot \sec^2 x) \, dx + C \)
Since \( \sin x \cdot \sec^2 x = \frac{\sin x}{\cos^2 x} = \sec x \tan x \):
\( y \cdot \sec^2 x = \int \sec x \tan x \, dx + C \)
\( y \cdot \sec^2 x = \sec x + C \)
Dividing both sides by \( \sec^2 x \):
\( y = \frac{\sec x}{\sec^2 x} + \frac{C}{\sec^2 x} \)
\( y = \cos x + C\cos^2 x \)
where \( C \) is an arbitrary constant.
In simple words: The integrating factor here is \( \sec^2 x \). When we multiply and integrate, the right side becomes the integral of \( \sec x \tan x \), which is simply \( \sec x \).

Exam Tip: Remember the trigonometric identity \( \sin x \sec^2 x = \sec x \tan x \) as it simplifies the integration step dramatically.

Question 2. Solve \( (1 + x)\frac{dy}{dx} - y = e^{3x}(x + 1)^2 \).
Answer:
The given differential equation is:
\( (1 + x)\frac{dy}{dx} - y = e^{3x}(x + 1)^2 \)
Dividing both sides by \( (1 + x) \):
\( \frac{dy}{dx} - \frac{1}{1+x}y = e^{3x}(1 + x) \) - (1)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where:
\( P = -\frac{1}{1+x} \) and \( Q = e^{3x}(1 + x) \)
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dx} = e^{\int -\frac{1}{1+x} \, dx} = e^{-\log|1+x|} = e^{\log(1+x)^{-1}} = \frac{1}{1+x} \)
The general solution is:
\( y \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dx + C \)
\( y \cdot \frac{1}{1+x} = \int \left[ e^{3x}(1 + x) \cdot \frac{1}{1+x} \right] \, dx + C \)
\( \frac{y}{1+x} = \int e^{3x} \, dx + C \)
\( \frac{y}{1+x} = \frac{e^{3x}}{3} + C \)
\( y = (1 + x) \left( \frac{e^{3x}}{3} + C \right) \)
where \( C \) is the constant of integration.
In simple words: Divide by \( 1+x \) to put it in linear form. The integrating factor is \( \frac{1}{1+x} \), which cancels the \( (1+x) \) factor on the right side and leaves a simple exponential integral.

Exam Tip: Simplifying the terms after multiplying by the Integrating Factor often leads to very easy integrations. Always check for factors that cancel out.

Question 3. Solve \( x\frac{dy}{dx} + y = x\log x \).
Answer:
The given differential equation is:
\( x\frac{dy}{dx} + y = x\log x \)
Dividing by \( x \):
\( \frac{dy}{dx} + \frac{1}{x}y = \log x \) - (1)
This is a linear differential equation where:
\( P = \frac{1}{x} \) and \( Q = \log x \)
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int \frac{1}{x} \, dx} = e^{\log x} = x \)
The general solution is:
\( y \cdot x = \int (x \log x) \, dx + C \) - (2)
We solve \( \int x \log x \, dx \) using integration by parts (taking \( \log x \) as the first function and \( x \) as the second function):
\( \int x \log x \, dx = \log x \left( \frac{x^2}{2} \right) - \int \left( \frac{1}{x} \cdot \frac{x^2}{2} \right) dx \)
\( = \frac{x^2}{2}\log x - \frac{1}{2}\int x \, dx \)
\( = \frac{x^2}{2}\log x - \frac{x^2}{4} \)
Substituting this back into (2):
\( xy = \frac{x^2}{2}\log x - \frac{x^2}{4} + C \)
Dividing both sides by \( x \):
\( y = \frac{x}{2}\log x - \frac{x}{4} + \frac{C}{x} \)
where \( C \) is the constant of integration.
In simple words: Divide by \( x \) to get the linear form, where the integrating factor is \( x \). Integrate the right side, \( x \log x \), using integration by parts, and then divide the final result by \( x \).

Exam Tip: Remember the ILATE rule for integration by parts, where logarithmic functions (L) always take priority over algebraic functions (A) as the first function \( u \).

Level III

Question 1. Solve \( \frac{dy}{dx} = \cos(x + y) \).
Answer:
The given differential equation is:
\( \frac{dy}{dx} = \cos(x + y) \) - (1)
Let \( x + y = v \). Differentiating both sides with respect to \( x \):
\( 1 + \frac{dy}{dx} = \frac{dv}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
Substituting these into equation (1):
\( \frac{dv}{dx} - 1 = \cos v \)
\( \frac{dv}{dx} = 1 + \cos v \)
Separating variables:
\( \frac{1}{1 + \cos v} \, dv = dx \) - (2)
We know that \( 1 + \cos v = 2\cos^2\left(\frac{v}{2}\right) \):
\( \frac{1}{2\cos^2\left(\frac{v}{2}\right)} \, dv = dx \)
\( \frac{1}{2}\sec^2\left(\frac{v}{2}\right) dv = dx \)
Integrating both sides:
\( \int \frac{1}{2}\sec^2\left(\frac{v}{2}\right) dv = \int dx \)
\( \tan\left(\frac{v}{2}\right) = x + C \)
Substitute \( v = x + y \) back:
\( \tan\left(\frac{x + y}{2}\right) = x + C \)
where \( C \) is the constant of integration.
In simple words: Substitute \( v = x + y \) to turn the trig expression into a single variable. This makes it separable, and we integrate using the half-angle formula for cosine.

Exam Tip: When the argument of a function is of the form \( ax + by + c \), substitute it with a new variable \( v \) to reduce the differential equation to a separable form.

Question 2. Solve \( y e^y \, dx = (y^3 + 2x e^y) dy \).
Answer:
The given equation can be rewritten as:
\( y e^y \frac{dx}{dy} = y^3 + 2x e^y \)
\( y e^y \frac{dx}{dy} - 2e^y x = y^3 \)
Dividing both sides by \( y e^y \):
\( \frac{dx}{dy} - \frac{2}{y}x = y^2 e^{-y} \) - (1)
This is a linear differential equation of the form \( \frac{dx}{dy} + Px = Q \), where \( x \) is the dependent variable and \( y \) is the independent variable:
\( P = -\frac{2}{y} \) and \( Q = y^2 e^{-y} \)
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dy} = e^{\int -\frac{2}{y} \, dy} = e^{-2\log y} = e^{\log(y^{-2})} = \frac{1}{y^2} \)
The general solution is:
\( x \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dy + C \)
\( x \cdot \frac{1}{y^2} = \int \left( y^2 e^{-y} \cdot \frac{1}{y^2} \right) dy + C \)
\( \frac{x}{y^2} = \int e^{-y} \, dy + C \)
\( \frac{x}{y^2} = -e^{-y} + C \)
\( x = y^2 (C - e^{-y}) \)
where \( C \) is an arbitrary constant.
In simple words: Rearrange the equation so that you can solve for \( \frac{dx}{dy} \). It is a linear equation in \( x \), so we calculate an integrating factor in terms of \( y \) and integrate to find the solution.

Exam Tip: Be ready for linear differential equations where \( x \) is the dependent variable (i.e., of the form \( \frac{dx}{dy} + Px = Q \)), which are solved in a perfectly analogous way to the standard form.

Question 3. Solve \( x^2 \frac{dy}{dx} = y(x + y) \).
Answer:
The given differential equation is:
\( \frac{dy}{dx} = \frac{xy + y^2}{x^2} \) - (1)
This is a homogeneous differential equation. Let \( y = vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting these into (1):
\( v + x\frac{dv}{dx} = \frac{x(vx) + (vx)^2}{x^2} = \frac{x^2(v + v^2)}{x^2} \)
\( v + x\frac{dv}{dx} = v + v^2 \)
\( x\frac{dv}{dx} = v^2 \)
Separating variables:
\( v^{-2} \, dv = \frac{dx}{x} \)
Integrating both sides:
\( \int v^{-2} \, dv = \int \frac{1}{x} \, dx \)
\( -\frac{1}{v} = \log|x| + C \)
Substitute \( v = \frac{y}{x} \):
\( -\frac{x}{y} = \log|x| + C \)
\( y = \frac{-x}{\log|x| + C} \)
This is the general solution.
In simple words: Since it is homogeneous, use \( y = vx \). This leaves a very simple separable equation \( x \frac{dv}{dx} = v^2 \), which is easily integrated.

Exam Tip: Homogeneous equations often simplify to extremely basic variable-separable equations. Don't worry if the intermediate steps look simple.

Question 4. Solve \( \frac{dy}{dx} + \frac{4x}{x^2 + 1}y = \frac{1}{(x^2 + 1)^3} \).
Answer:
The given equation is already in the standard form \( \frac{dy}{dx} + Py = Q \), where:
\( P = \frac{4x}{x^2 + 1} \) and \( Q = \frac{1}{(x^2 + 1)^3} \)
First, find the Integrating Factor (I.F.):
\( \text{I.F.} = e^{\int P \, dx} = e^{\int \frac{4x}{x^2 + 1} \, dx} \)
For the integral, let \( x^2 + 1 = u \implies 2x \, dx = du \implies 4x \, dx = 2 \, du \):
\( \int \frac{4x}{x^2 + 1} \, dx = 2 \int \frac{1}{u} \, du = 2\log(x^2 + 1) = \log(x^2 + 1)^2 \)
Thus:
\( \text{I.F.} = e^{\log(x^2 + 1)^2} = (x^2 + 1)^2 \)
The general solution is:
\( y \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dx + C \)
\( y \cdot (x^2 + 1)^2 = \int \left[ \frac{1}{(x^2 + 1)^3} \cdot (x^2 + 1)^2 \right] \, dx + C \)
\( y \cdot (x^2 + 1)^2 = \int \frac{1}{x^2 + 1} \, dx + C \)
\( y \cdot (x^2 + 1)^2 = \tan^{-1} x + C \)
\( y = \frac{\tan^{-1} x + C}{(x^2 + 1)^2} \)
where \( C \) is an arbitrary constant.
In simple words: The integrating factor works out to be \( (x^2 + 1)^2 \). Multiplying this by the right side simplifies it to \( \frac{1}{x^2 + 1} \), which integrates to \( \tan^{-1} x \).

Exam Tip: Be careful when simplifying powers of logarithms: \( e^{a \log f(x)} = f(x)^a \). This is a standard simplification step in finding Integrating Factors.

Question 5. Solve the differential equation \( (x + 2y^2)\frac{dy}{dx} = y \), given that \( y = 1 \) when \( x = 2 \).
Answer:
The given equation is:
\( (x + 2y^2)\frac{dy}{dx} = y \)
Taking reciprocal on both sides:
\( \frac{dx}{dy} = \frac{x + 2y^2}{y} \)
\( \frac{dx}{dy} = \frac{x}{y} + 2y \)
\( \frac{dx}{dy} - \frac{1}{y}x = 2y \) - (1)
This is a linear differential equation in \( x \), where:
\( P = -\frac{1}{y} \) and \( Q = 2y \)
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\log|y|} = \frac{1}{y} \)
The general solution is:
\( x \cdot (\text{I.F.}) = \int [Q \cdot (\text{I.F.})] \, dy + C \)
\( x \cdot \frac{1}{y} = \int \left( 2y \cdot \frac{1}{y} \right) dy + C \)
\( \frac{x}{y} = \int 2 \, dy + C \)
\( \frac{x}{y} = 2y + C \)
\( x = 2y^2 + Cy \) - (2)
Now, apply the initial condition \( y = 1 \) when \( x = 2 \):
\( 2 = 2(1)^2 + C(1) \)
\( 2 = 2 + C \implies C = 0 \)
Substituting \( C = 0 \) into equation (2):
\( x = 2y^2 \)
This is the required particular solution.
In simple words: Turn the equation upside down to make it linear in \( x \). Solve it to get \( x = 2y^2 + Cy \). Using the starting values \( x=2 \) and \( y=1 \), we find that \( C=0 \), so the final answer is \( x = 2y^2 \).

Exam Tip: When a differential equation has a single term of \( y \) and multiple terms of \( x \), it is often beneficial to write it as \( \frac{dx}{dy} + P(y)x = Q(y) \).

Questions for Self Evaluation

Question 1. Write the order and degree of the differential equation \( \left(\frac{d^3 y}{dx^3}\right)^2 + \frac{d^2 y}{dx^2} + \sin\left(\frac{dy}{dx}\right) = 0 \).
Answer:
The given differential equation is:
\( \left(\frac{d^3 y}{dx^3}\right)^2 + \frac{d^2 y}{dx^2} + \sin\left(\frac{dy}{dx}\right) = 0 \)
Here, the highest-order derivative present is \( \frac{d^3 y}{dx^3} \), so the order of the differential equation is 3.
However, because of the term \( \sin\left(\frac{dy}{dx}\right) \), the equation cannot be written as a polynomial in its derivatives.
Therefore, the degree of this differential equation is not defined.
In simple words: The highest derivative is the third derivative, so the order is 3. Since a derivative is inside a sine function, we cannot define the degree.

Exam Tip: If any derivative is present as an argument of a trigonometric, exponential, or logarithmic function (e.g., \( \sin(y') \), \( e^{y''} \)), the degree of the equation is always not defined.

Question 2. Form the differential equation representing the family of ellipses having foci on the x-axis and centre at the origin.
Answer:
The equation of the family of ellipses with center at the origin and foci on the x-axis is:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) - (1)
where \( a \) and \( b \) are arbitrary constants.
Differentiating equation (1) with respect to \( x \):
\( \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \)
\( \frac{x}{a^2} + \frac{y}{b^2}\frac{dy}{dx} = 0 \)
\( \frac{y}{x}\frac{dy}{dx} = -\frac{b^2}{a^2} \) - (2)
Differentiating both sides of equation (2) with respect to \( x \) using the product/quotient rule:
\( \frac{d}{dx} \left[ \frac{y \frac{dy}{dx}}{x} \right] = 0 \)
\( \frac{x \frac{d}{dx}\left(y \frac{dy}{dx}\right) - y \frac{dy}{dx} \cdot 1}{x^2} = 0 \)
Using the product rule on \( y \frac{dy}{dx} \):
\( x \left[ y \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 \right] - y \frac{dy}{dx} = 0 \)
\( xy\frac{d^2 y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0 \)
This is the required differential equation.
In simple words: Write the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Differentiate it once to separate the constants on one side, then differentiate again to eliminate them entirely.

Exam Tip: Moving all arbitrary constants to one side before the final differentiation is a highly efficient way to eliminate them without complex substitution.

Question 3. Solve the differential equation: \( (\tan^{-1} y - x) \, dy = (1 + y^2) \, dx \), given that \( y = 0 \) when \( x = 0 \).
Answer:
The given equation can be rewritten as:
\( (1 + y^2)\frac{dx}{dy} = \tan^{-1} y - x \)
\( \frac{dx}{dy} + \frac{1}{1 + y^2}x = \frac{\tan^{-1} y}{1 + y^2} \) - (1)
This is a linear differential equation in \( x \), where:
\( P = \frac{1}{1+y^2} \) and \( Q = \frac{\tan^{-1} y}{1+y^2} \)
The Integrating Factor (I.F.) is:
\( \text{I.F.} = e^{\int P \, dy} = e^{\int \frac{1}{1+y^2} \, dy} = e^{\tan^{-1} y} \)
The general solution is:
\( x \cdot e^{\tan^{-1} y} = \int \left( \frac{\tan^{-1} y}{1+y^2} \cdot e^{\tan^{-1} y} \right) dy + C \) - (2)
For the RHS integral, let \( \tan^{-1} y = t \implies \frac{1}{1+y^2} \, dy = dt \):
\( \int t e^t \, dt = t e^t - e^t = e^t(t - 1) = e^{\tan^{-1} y}(\tan^{-1} y - 1) \)
Substituting this back into (2):
\( x e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + C \)
Dividing by \( e^{\tan^{-1} y} \):
\( x = \tan^{-1} y - 1 + C e^{-\tan^{-1} y} \) - (3)
Now, apply the initial condition \( y = 0 \) when \( x = 0 \):
\( 0 = \tan^{-1} 0 - 1 + C e^{-\tan^{-1} 0} \)
\( 0 = 0 - 1 + C \implies C = 1 \)
Substituting \( C = 1 \) into equation (3):
\( x = \tan^{-1} y - 1 + e^{-\tan^{-1} y} \)
This is the required particular solution.
In simple words: Rearrange the equation into linear form with \( x \) as the dependent variable. Calculate the integrating factor \( e^{\tan^{-1} y} \), solve the integral on the right using substitution, and then find the constant \( C = 1 \) using the initial condition.

Exam Tip: Standard integrals like \( \int t e^t \, dt = e^t(t-1) \) are extremely common in linear differential equations. Memorizing them saves valuable exam time.

Question 4. Solve the differential equation: \( x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx \).
Answer:
The given differential equation is:
\( x \, dy = \left(y + \sqrt{x^2 + y^2}\right) dx \)
\( \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x} \) - (1)
Let \( y = vx \), then \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute these into equation (1):
\( v + x\frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2 x^2}}{x} \)
\( v + x\frac{dv}{dx} = v + \sqrt{1 + v^2} \)
\( x\frac{dv}{dx} = \sqrt{1 + v^2} \)
Separating variables:
\( \frac{1}{\sqrt{1 + v^2}} \, dv = \frac{dx}{x} \)
Integrating both sides:
\( \int \frac{1}{\sqrt{1 + v^2}} \, dv = \int \frac{1}{x} \, dx \)
\( \log|v + \sqrt{1 + v^2}| = \log|x| + \log C \)
\( v + \sqrt{1 + v^2} = Cx \)
Substituting back \( v = \frac{y}{x} \):
\( \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx \)
\( \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = Cx \)
\( y + \sqrt{x^2 + y^2} = Cx^2 \)
where \( C \) is an arbitrary constant.
In simple words: This homogeneous equation is solved by letting \( y = vx \). This simplifies it to a variable-separable equation, which we integrate using the standard log-based formula.

Exam Tip: Be sure to write the constant of integration in logarithmic form \( \log C \) when both sides of the integrated equation result in natural logarithms, as it makes simplification much cleaner.

Question 5. Solve the differential equation: \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \).
Answer:
The given differential equation is:
\( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \)
Dividing both sides by \( x \log x \):
\( \frac{dy}{dx} + \frac{1}{x \log x}y = \frac{2}{x^2} \) - (1)
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), where:
\( P = \frac{1}{x \log x} \) and \( Q = \frac{2}{x^2} \)
First, find the Integrating Factor (I.F.):
\( \text{I.F.} = e^{\int P \, dx} = e^{\int \frac{1}{x} \log x \, dx} \)
For the integral, let \( \log x = t \implies \frac{1}{x} \, dx = dt \):
\( \int \frac{1}{x \log x} \, dx = \int \frac{1}{t} \, dt = \log|t| = \log|\log x| \)
Thus:
\( \text{I.F.} = e^{\log(\log x)} = \log x \)
The general solution is:
\( y \cdot \log x = \int \left( \frac{2}{x^2} \log x \right) dx + C \) - (2)
Let us solve the integral on the RHS, \( 2 \int x^{-2} \log x \, dx \), using integration by parts (with \( u = \log x \) and \( dv = x^{-2} \, dx \)):
\( du = \frac{1}{x} \, dx \), \( v = -\frac{1}{x} \)
\( \int x^{-2} \log x \, dx = (\log x)\left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right)\left(\frac{1}{x}\right) dx \)
\( = -\frac{\log x}{x} + \int x^{-2} \, dx \)
\( = -\frac{\log x}{x} - \frac{1}{x} = -\frac{1 + \log x}{x} \)
Substitute this back into (2):
\( y \log x = -2 \left( \frac{1 + \log x}{x} \right) + C \)
This is the general solution.
In simple words: Divide by \( x \log x \) to make it linear. The integrating factor is \( \log x \). Multiplying and integrating by parts on the right side gives the final answer.

Exam Tip: Be sure to keep track of negative signs during Integration by Parts, especially when the integration of \( v \) introduces a minus sign.

Question 6. Solve the differential equation: \( x^2 \, dy + (y^2 + xy) \, dx = 0 \), given that \( y(1) = 1 \).
Answer:
The given differential equation can be written as:
\( x^2 \frac{dy}{dx} = -(y^2 + xy) \)
\( \frac{dy}{dx} = -\frac{xy + y^2}{x^2} \) - (1)
This is a homogeneous differential equation. Let \( y = vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting these into (1):
\( v + x\frac{dv}{dx} = -\frac{x(vx) + (vx)^2}{x^2} = -(v + v^2) \)
\( x\frac{dv}{dx} = -2v - v^2 = -v(v + 2) \)
Separating variables:
\( \frac{1}{v(v + 2)} \, dv = -\frac{dx}{x} \) - (2)
Using partial fractions on the LHS:
\( \frac{1}{v(v + 2)} = \frac{1}{2} \left( \frac{1}{v} - \frac{1}{v+2} \right) \)
Integrating both sides of (2):
\( \frac{1}{2} \int \left( \frac{1}{v} - \frac{1}{v+2} \right) dv = -\int \frac{1}{x} \, dx \)
\( \frac{1}{2} \left( \log|v| - \log|v+2| \right) = -\log|x| + C' \)
Multiplying by 2:
\( \log\left| \frac{v}{v+2} \right| = -2\log|x| + 2C' \)
\( \log\left| \frac{v}{v+2} \right| = \log\left| \frac{1}{x^2} \right| + \log C \)
\( \frac{v}{v+2} = \frac{C}{x^2} \)
Substituting back \( v = \frac{y}{x} \):
\( \frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C}{x^2} \)
\( \frac{y}{y + 2x} = \frac{C}{x^2} \) - (3)
Apply the initial condition \( y = 1 \) when \( x = 1 \):
\( \frac{1}{1 + 2(1)} = \frac{C}{1^2} \implies C = \frac{1}{3} \)
Substituting \( C = \frac{1}{3} \) back into (3):
\( \frac{y}{y + 2x} = \frac{1}{3x^2} \)
\( 3x^2 y = y + 2x \)
\( y(3x^2 - 1) = 2x \)
\( y = \frac{2x}{3x^2 - 1} \)
This is the particular solution.
In simple words: Since the equation is homogeneous, let \( y = vx \). The variables then separate easily, and we integrate using partial fractions. After finding the general solution, substitute \( x=1, y=1 \) to find that the constant \( C = 1/3 \).

Exam Tip: Partial fraction decomposition is a very common tool in integrating separated variables. Practice decomposing simple linear factors like \( \frac{1}{v(v+a)} \).

Question 7. Show that the differential equation \( 2y e^{x/y} \, dx + (y - 2x e^{x/y}) dy = 0 \) is homogeneous and find its particular solution given that \( y = 1 \) when \( x = 0 \).
Answer:
The given differential equation can be written as:
\( 2y e^{x/y} \frac{dx}{dy} + y - 2x e^{x/y} = 0 \)
\( \frac{dx}{dy} = \frac{2x e^{x/y} - y}{2y e^{x/y}} \) - (1)
Let \( f(x, y) = \frac{2x e^{x/y} - y}{2y e^{x/y}} \). Replacing \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\( f(\lambda x, \lambda y) = \frac{2(\lambda x) e^{\lambda x / \lambda y} - \lambda y}{2(\lambda y) e^{\lambda x / \lambda y}} = \frac{\lambda (2x e^{x/y} - y)}{\lambda (2y e^{x/y})} = f(x, y) \)
Thus, the differential equation is homogeneous of degree 0.
Since the equation contains terms of \( \frac{x}{y} \), we substitute \( x = vy \). Differentiating both sides with respect to \( y \):
\( \frac{dx}{dy} = v + y\frac{dv}{dy} \)
Substituting these into (1):
\( v + y\frac{dv}{dy} = \frac{2(vy) e^v - y}{2y e^v} = \frac{y(2v e^v - 1)}{2y e^v} \)
\( v + y\frac{dv}{dy} = \frac{2v e^v - 1}{2 e^v} \)
\( y\frac{dv}{dy} = \frac{2v e^v - 1}{2 e^v} - v = \frac{2v e^v - 1 - 2v e^v}{2 e^v} = -\frac{1}{2e^v} \)
Separating variables:
\( 2 e^v \, dv = -\frac{dy}{y} \)
Integrating both sides:
\( \int 2 e^v \, dv = -\int \frac{1}{y} \, dy \)
\( 2 e^v = -\log|y| + C \) - (2)
Substitute \( v = \frac{x}{y} \) back:
\( 2 e^{x/y} + \log|y| = C \) - (3)
Now, apply the initial condition \( y = 1 \) when \( x = 0 \) (note that \( y(0) = 1 \) means \( x = 0, y = 1 \)):
\( 2 e^{0/1} + \log|1| = C \)
\( 2(1) + 0 = C \implies C = 2 \)
Substituting \( C = 2 \) into (3):
\( 2 e^{x/y} + \log|y| = 2 \)
This is the required particular solution.
In simple words: Show the equation does not change when substituting \( \lambda x \) and \( \lambda y \). Since it contains \( x/y \), substitute \( x = vy \) instead of the usual \( y = vx \), separate variables, integrate, and find \( C = 2 \) using the given values.

Exam Tip: When a homogeneous differential equation has terms with \( \frac{x}{y} \), using the substitution \( x = vy \) and finding \( \frac{dx}{dy} \) makes the algebraic simplification much easier than using \( y = vx \).

Question 8. Find the particular solution of differential equation \( \frac{dy}{dx} + y \cot x = 2x + x^2 \cot x \), given that \( y\left(\frac{\pi}{2}\right) = 0 \).
Answer:
The given differential equation is in the standard linear form \( \frac{dy}{dx} + Py = Q \), where:
\( P = \cot x \) and \( Q = 2x + x^2 \cot x \)
First, find the Integrating Factor (I.F.):
\( \text{I.F.} = e^{\int P \, dx} = e^{\int \cot x \, dx} = e^{\log| \sin x |} = \sin x \)
The general solution is:
\( y \cdot \sin x = \int [ (2x + x^2 \cot x) \sin x ] \, dx + C \) - (1)
Let us simplify the integrand:
\( (2x + x^2 \cot x) \sin x = 2x \sin x + x^2 \left(\frac{\cos x}{\sin x}\right) \sin x = 2x \sin x + x^2 \cos x \)
Notice that this integrand is the exact derivative of \( x^2 \sin x \) using the product rule:
\( \frac{d}{dx} (x^2 \sin x) = 2x \sin x + x^2 \cos x \)
Therefore:
\( \int (2x \sin x + x^2 \cos x) \, dx = x^2 \sin x \)
Substituting this back into (1):
\( y \sin x = x^2 \sin x + C \) - (2)
Now, apply the initial condition \( y\left(\frac{\pi}{2}\right) = 0 \) (i.e., \( y = 0 \) when \( x = \frac{\pi}{2} \)):
\( 0 \cdot \sin\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + C \)
\( 0 = \frac{\pi^2}{4} (1) + C \implies C = -\frac{\pi^2}{4} \)
Substitute \( C = -\frac{\pi^2}{4} \) back into (2):
\( y \sin x = x^2 \sin x - \frac{\pi^2}{4} \)
Dividing by \( \sin x \):
\( y = x^2 - \frac{\pi^2}{4 \sin x} \)
\( y = x^2 - \frac{\pi^2}{4}\csc x \)
This is the required particular solution.
In simple words: The integrating factor is \( \sin x \). When we multiply the right-hand side by \( \sin x \), it simplifies perfectly to the exact derivative of \( x^2 \sin x \). Integrating it and using the initial condition yields the constant \( C = -\frac{\pi^2}{4} \).

Exam Tip: Recognizing exact derivatives of products like \( \frac{d}{dx}(x^2 \sin x) = 2x\sin x + x^2\cos x \) saves you from doing long integration by parts twice. Always look for such patterns on competitive exams!

Question. Solve the D.E : \( x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x \).
Answer: Divide by \( x \log x \)
\( \frac{dy}{dx} + \frac{1}{x \log x} y = \frac{2}{x^2} \)
Comparing with \( \frac{dy}{dx} + Py = \theta \)
We have, \( P = \frac{1}{x \log x} \) & \( \theta = \frac{2}{x^2} \)
\( I.F. = e^{\int P dx} = e^{\int \frac{1}{x \log x} dx} \)
Put \( \log x = t \Rightarrow \frac{1}{x} dx = dt \)
\( \therefore I.F. = e^{\int \frac{dt}{t}} = e^{\log t} = t = \log x \)
\( \therefore I.F. = \log x \)
Solution is given by
\( y.(I.F.) = \int \theta.(I.F.) dx + C \)
\( \Rightarrow y \log x = 2 \int \frac{1}{x^2} \cdot \log x dx + C \)
\( \Rightarrow y \log x = 2 \left[ \log x \left( -\frac{1}{x} \right) - \int \frac{1}{x} \left( -\frac{1}{x} \right) dx \right] + C \)
\( \Rightarrow y \log x = 2 \left[ -\frac{\log x}{x} + \int \frac{1}{x^2} dx \right] + C \)
\( \Rightarrow y \log x = 2 \left[ -\frac{\log x}{x} - \frac{1}{x} \right] + C \)
\( \Rightarrow y \log x = -\frac{2}{x}(\log x + 1) + C \) is the required solution.

Question. Solve the D.E. : \( x \frac{dy}{dx} + y - x + xy \cot x = 0 \).
Answer: We have, \( x \frac{dy}{dx} + y(1 + x \cot x) = x \)
Divide by \( x \)
\( \Rightarrow \frac{dy}{dx} + y \left( \frac{1}{x} + \cot x \right) = 1 \)
Comparing with \( \frac{dy}{dx} + Py = \theta \)
We have, \( P = \frac{1}{x} + \cot x \) and \( \theta = 1 \)
\( I.F. = e^{\int \left( \frac{1}{x} + \cot x \right) dx} = e^{\log x \log(\sin x)} = e^{\log(x \sin x)} = x \sin x \)
\( \therefore I.F. = x \sin x \)
Solution is given by
\( \Rightarrow y(I.F.) = \int \theta (I.F.) dx + C \)
\( \Rightarrow y (x \sin x) = \int 1.(x \sin x) dx + C \)
\( \Rightarrow y(x \sin x) = x(-\cos x) - \int (1).(-\cos x) dx + C \)
\( \Rightarrow y(x \sin x) = -x \cos x + \sin x + C \) ans.

Question. Solve the initial value problem \( (x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1) \cos x ; y(0) = 0 \).
Answer: We have, \( (x^2 + 1)\frac{dy}{dx} - 2xy = (x^2 + 1)^2 \cdot \cos x \)
Divide by \( x^2 + 1 \)
\( \Rightarrow \frac{dy}{dx} - \frac{2x}{1+x^2} y = (x^2 + 1) \cdot \cos x \)
Comparing with \( \frac{dy}{dx} + Py = \theta \)
We have, \( P = \frac{-2x}{x^2+1} \) and \( \theta = (x^2 + 1) \cos x \)
\( I.F. = e^{\int P dx} = e^{-\int \frac{2x}{1+x^2} dx} \)
Put \( 1 + x^2 = t \Rightarrow 2x dx = dt \)
\( \therefore I.F. = e^{-\int \frac{dt}{t}} = e^{-\log t} = e^{\log(t)^{-1}} \)
\( I.F. = \frac{1}{t} = \frac{1}{1+x^2} \)
\( \therefore I.F. = \frac{1}{1+x^2} \)
Now solution is given by \( y(I.F.) = \int \theta \cdot (I.F.) dx + C \)
\( \Rightarrow y \times \frac{1}{x^2+1} = \int (x^2 + 1) \cdot \cos x \cdot \frac{1}{1+x^2} dx + C \)
\( \Rightarrow \frac{y}{x^2+1} = \sin x + C \)
Initial condition \( y(0) = 0 \Rightarrow x = 0 \) and \( y = 0 \)
\( \Rightarrow 0 = \sin(0) = C \)
\( \Rightarrow C = 0 \)
\( \therefore \frac{y}{x^2 + 1} = \sin x \)
\( \Rightarrow y = (x^2 + 1) \sin x \) is the required particular solution.

Question. Solve the D.E. \( \frac{dy}{dx} = \frac{-x+y \cos x}{1+\sin x} \).
Answer: We have, \( \frac{dy}{dx} = \frac{-x+y \cos x}{1+\sin x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{-x}{1+\sin x} - \frac{y \cos x}{1+\sin x} \)
\( \Rightarrow \frac{dy}{dx} + \frac{y \cos x}{1+\sin x} = \frac{-x}{1+\sin x} \)
Comparing with \( \frac{dy}{dx} + Py = \theta \)
We have, \( P = \frac{\cos x}{1+\sin x} \) and \( \theta = \frac{-x}{1+\sin x} \)
\( I.F. = e^{\int \frac{\cos x}{1+\sin x} dx} \)
Put \( 1 + \sin x = t \Rightarrow \cos x dx = dt \)
\( \therefore I.F. = e^{\int \frac{dt}{t}} = e^{\log t} = t = 1 + \sin x \)
\( \therefore I.F. = 1 + \sin x \)
Solution is given by \( y(I.F.) = \int \theta \cdot (I.F.) dx + C \)
\( \Rightarrow y(1 + \sin x) = \int \frac{-x}{(1+\sin x)} \times (1 + \sin x) dx + C \)
\( \Rightarrow y(1 + \sin x) = \frac{(-x)^2}{2} + C \) is the required solution.

Question. Find the general solution \( \frac{dy}{dx} - 2y = \cos(3x) \).
Answer: Comparing with \( \frac{dy}{dx} + Py = \theta \)
We have, \( P = -2 \) and \( \theta = \cos(3x) \)
\( I.F. = e^{\int -2dx} \)
\( I.F. = e^{-2x} \)
Solution is given by \( y(I.F.) = \int \theta (I.F.) dx + C \)
\( y \cdot e^{-2x} = \int e^{-2x} \cos(3x) dx + C \)
\( \Rightarrow y \cdot e^{-2x} = I + C \) ............. (i)
Where \( I = \int e^{-2x} \cos(3x) dx \)
\( I = \cos(3x) \cdot \frac{e^{-2x}}{-2} - \int -3 \sin(3x) \cdot \frac{e^{-2x}}{-2} dx \)
\( I = -\frac{1}{2} \cos(3x) \cdot e^{-2x} - \frac{3}{2} \int e^{-2x} \cdot \sin(3x) dx \)
\( = -\frac{1}{2} \cdot \cos(3x) e^{-2x} - \frac{3}{2} \left[ \sin(3x) \cdot \frac{e^{-2x}}{-2} - \int 3 \cos(3x) \cdot \frac{e^{-2x}}{-2} dx \right] \)
\( = -\frac{1}{2} \cos(3x) e^{-2x} - \frac{3}{2} \left[ -\frac{1}{2} \sin(3x) e^{-2x} + \frac{3}{2} I \right] \)
\( I = -\frac{1}{2} \cdot e^{-2x} \cdot \cos(3x) + \frac{3}{4} e^{-2x} \cdot \sin(3x) - \frac{9}{4} I \)
\( \Rightarrow I + \frac{9}{4} I = \frac{e^{-2x}}{4} (-2 \cos(3x) + 3 \sin(3x)) \)
\( \Rightarrow 13I = e^{-2x} (-2 \cos(3x) + 3 \sin(3x)) \)
\( \Rightarrow I = \frac{e^{-2x}}{13} (-2 \cos(3x) + 3 \sin(3x)) \)
\( \therefore \) equation (i) become
\( y \cdot e^{-2x} = \frac{e^{-2x}}{13} (-2 \cos(3x) + 3 \sin(3x)) + C \)
\( \Rightarrow y = \frac{1}{13} (-2 \cos(3x) + 3 \sin(3x)) + C e^{2x} \) ans.

Question. Solve the D.E. \( \cos^2 x \frac{dy}{dx} + y = \tan x \).
Answer: Divide by \( \cos^2 x \)
\( \Rightarrow \frac{dy}{dx} + y \sec^2 x = \tan x \cdot \sec^2 x \)
Here \( P = \sec^2 x \) and \( \theta = \tan x \sec^2 x \)
\( I.F. = e^{\int P dx} = e^{\int \sec^2 x dx} = e^{\tan x} \)
\( \dots I.F. = e^{\tan x} \)
Solving is given by
\( y.(I.F.) = \int \theta (I.F.) dx + C \)
\( \Rightarrow y \cdot e^{\tan x} = \int \tan x \cdot \sec^2 x \cdot e^{\tan x} dx + C \)
Put \( \tan x = t \)
\( \Rightarrow \sec^2 x dx = dt \)
\( \therefore y \cdot e^{\tan x} = \int e^t \cdot t dt + C \)
\( \Rightarrow y e^{\tan x} = t \cdot e^t - \int e^t dt + C \)
\( \Rightarrow y e^{\tan x} = t e^t - e^t + C \)
\( \Rightarrow y e^{\tan x} = e^{\tan x} (\tan x - 1) + C \) ans.

Question. Solve the D.E. \( (x^2 + 1)\frac{dy}{dx} + 2xy = \sqrt{x^2 + 4} \).
Answer: Divide by \( (x^2 + 1) \)
\( \frac{dy}{dx} + \frac{2x}{x^2+1} y = \frac{\sqrt{x^2+4}}{x^2+1} \)
Comparing with \( \frac{dy}{dx} + Py = \theta \)
We have, \( P = \frac{2x}{x^2+1} \) and \( \theta = \frac{\sqrt{x^2+4}}{x^2+1} \)
\( I.F. = e^{\int P dx} = e^{\int \left( \frac{2x}{x^2+1} \right) dx} \)
Put \( x^2 + 1 = t \Rightarrow 2x dx = dt \)
\( I.F. = e^{\int \frac{dt}{t}} = e^{\log t} = t = x^2 + 1 \)
\( \therefore I.F. = x^2 + 1 \)
Solution is given by
\( y.(I.F.) = \int \theta (I.F.) dx + C \)
\( \Rightarrow y \cdot (x^2 + 1) = \int \frac{\sqrt{x^2+4}}{x^2+1} \cdot (x^2 + 1) dx + C \)
\( \Rightarrow y(x^2 + 1) = \int \sqrt{x^2 + 4} dx + C \)
\( \Rightarrow y(x^2 + 1) = \frac{x}{2}\sqrt{x^2+4} + \frac{4}{2}\log|x + \sqrt{x^2+4}| + C \)
\( \Rightarrow y(x^2 + 1) = \frac{x}{2}\sqrt{x^2+4} + 2\log|x + \sqrt{x^2+4}| + C \)

Question. Find the particular solution of the D.E. \( \frac{dy}{dx} + y = \cos x - \sin x \) given \( y(0) = 2 \).
Answer: Comparing with \( \frac{dy}{dx} + Py = \theta \)
Where \( P = 1 \) and \( \theta = \cos x - \sin x \)
\( I.F. = e^{\int P dx} = e^{\int 1.dx} = e^x \Rightarrow I.F. = e^x \)
Solution is given by
\( y.(I.F.) = \int \theta (I.F.) dx + C \)
\( \Rightarrow y e^x = \int e^x \cdot (\cos x - \sin x) dx + C \)
\( \Rightarrow y e^x = \int e^x \cdot \cos x dx - \int e^x \sin x dx + C \)
\( \Rightarrow y e^x = \cos x e^x - \int -\sin x \cdot e^x - \int e^x \sin x dx - \int e^x \sin x dx + C \)
\( \Rightarrow y e^x = e^x \cos x + \int e^x \sin x dx - \int e^x \sin x dx + C \)
\( y e^x = e^x \cos x + C \)
Put in Initial condition i.e., \( x = 0 \) and \( y = 2 \)
\( \Rightarrow 2 e^0 = e^0 \cdot \cos 0 + C \)
\( \Rightarrow 2 = C \)
\( \therefore y e^x = e^x \cos x + 2 \)
(or) \( y = \cos x + 2 e^x \) is the particular solution

Question. Find the general solution of the \( \frac{dy}{dx} + x \sin(2y) = x^3 \cos^2 y \).
Answer: We have \( \frac{dy}{dx} + x \sin(2y) = x^3 \cos^2 y \)
Divide by \( \cos^2 y \)
\( \Rightarrow \sec^2 y \frac{dy}{dx} + x \sin(2y) = x^3 \)
\( \Rightarrow \sec^2 y \frac{dy}{dx} + x \cdot \frac{2 \sin y \cos y}{\cos^2 y} = x^3 \)
\( \Rightarrow \sec^2 y \frac{dy}{dx} + 2x \tan y = x^3 \)
Let \( \tan y = V \Rightarrow \sec^2 y \frac{dy}{dx} = \frac{dV}{dx} \)
\( \therefore \frac{dV}{dx} + 2xV = x^3 \)
This a linear D.E of the form \( \frac{dV}{dx} + pV = Q \)
Here \( p = 2x \) and

Question. Find the particular solution of the D.E. \( \frac{dy}{dx} + y = \cos x - \sin x \) given \( y(0) = 2 \).
Answer: Compare with \( \frac{dy}{dx} + Py = \theta \).
We have, \( P = 1 \) and \( \theta = \cos x - \sin x \)
\( I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x \Rightarrow I.F. = e^x \)
Solution is given by
\( y.(I.F.) = \int \theta (I.F.) dx + C \)
\( \Rightarrow y e^x = \int e^x \cdot (\cos x \cdot \sin x) dx + C \)
\( \Rightarrow y e^x = \int e^x \cos x dx - \int e^x \sin x dx + C \)
\( \Rightarrow y e^x = \cos x e^x - \sin x dx + C \)
\( \Rightarrow y e^x = \cos x + \int e^x \sin x dx - \int e^x \sin x dx + C \)
\( \Rightarrow y e^x = e^x \cos x + C \)
Put initial condition i.e., \( x = 0 \) & \( y = 2 \)
\( \Rightarrow 2 e^0 = e^0 \cdot \cos 0 + C \)
\( \Rightarrow 2 = C \)
\( \therefore y e^x = e^x \cdot \cos x + 2 \)
Or \( y = \cos x + 2 e^{-x} \) is the particular solution ans.

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