CBSE Class 12 Mathematics Linear Differential Equations Worksheet Set 06

CBSE Class 12 Mathematics Linear Differential Equations (6). 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.

Formation of Differential Equations

Question 1. From the D.E. of the family of curves represented by \( c(y + c)^2 = x^3 \) where \( c \) is a parameter.
Answer: We start with the given equation: \[ c(y + c)^2 = x^3 \quad \dots (i) \] Differentiating both sides with respect to \( x \): \[ 2c(y + c)\frac{dy}{dx} = 3x^2 \quad \dots (ii) \] Dividing equation (i) by (ii): \[ \frac{y+c}{2 \frac{dy}{dx}} = \frac{x}{3} \]
\( \implies y + c = \frac{2x}{3} \frac{dy}{dx} \)
\( \implies c = \frac{2x}{3} \frac{dy}{dx} - y \) Substituting the expressions for \( y + c \) and \( c \) back into equation (i): \[ \left( \frac{2x}{3} \frac{dy}{dx} - y \right) \left( \frac{2x}{3} \frac{dy}{dx} \right)^2 = x^3 \]
\( \implies \left( \frac{2x}{3} \frac{dy}{dx} - y \right) \left( \frac{4x^2}{9} \left( \frac{dy}{dx} \right)^2 \right) = x^3 \)
\( \implies \frac{4}{9} \left( \frac{2x}{3} \frac{dy}{dx} - y \right) \left( \frac{dy}{dx} \right)^2 = x \)
\( \implies \frac{8x}{27} \left( \frac{dy}{dx} \right)^3 - \frac{4y}{9} \left( \frac{dy}{dx} \right)^2 = x \)
\( \implies 8x \left( \frac{dy}{dx} \right)^3 - 12y \left( \frac{dy}{dx} \right)^2 = 27x \) which represents the required differential equation.
In simple words: Eliminate the arbitrary constant \( c \) by differentiating and substituting its value back into the original curve equation.

Exam Tip: When forming a differential equation, the order of the differential equation must equal the number of independent arbitrary constants in the family of curves.

Question 2. Form the D.E. corresponding to \( y^2 = a(b - x)(b + x) \) where \( a \) & \( b \) are parameters.
Answer: The given equation of the family of curves is: \[ y^2 = a(b^2 - x^2) \quad \dots (i) \] Differentiating both sides with respect to \( x \): \[ 2y \frac{dy}{dx} = -2ax \]
\( \implies y \frac{dy}{dx} = -ax \quad \dots (ii) \) Differentiating once more with respect to \( x \): \[ y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = -a \] Substituting the value of \( -a = \frac{y}{x} \frac{dy}{dx} \) from equation (ii): \[ y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = \frac{y}{x} \frac{dy}{dx} \]
\( \implies xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0 \) This simplifies to the required differential equation.
In simple words: Differentiate the equation twice to eliminate both parameters \( a \) and \( b \). Replace the constants using the relation obtained from the first derivative.

Exam Tip: Since there are two parameters (\( a \) and \( b \)), we must differentiate twice to obtain a second-order differential equation.

Question 3. Form the D.E. of the family of curve \( y = ae^{bx} \) where \( a \) & \( b \) are parameter.
Answer: Let the given equation of the curve be: \[ y = ae^{bx} \quad \dots (i) \] Differentiating with respect to \( x \): \[ \frac{dy}{dx} = bae^{bx} \]
\( \implies \frac{dy}{dx} = by \quad \dots (ii) \) Differentiating again with respect to \( x \): \[ \frac{d^2y}{dx^2} = b \frac{dy}{dx} \quad \dots (iii) \] From equation (ii), we can write \( b = \frac{1}{y} \frac{dy}{dx} \). Substituting this value of \( b \) in (iii): \[ \frac{d^2y}{dx^2} = \left( \frac{1}{y} \frac{dy}{dx} \right) \frac{dy}{dx} \]
\( \implies y \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2 \)
\( \implies y \frac{d^2y}{dx^2} - \left( \frac{dy}{dx} \right)^2 = 0 \) This yields the required differential equation.
In simple words: Eliminate the constants \( a \) and \( b \) by finding the first and second derivatives, then expressing \( b \) in terms of \( y \) and its derivative to substitute it.

Exam Tip: For exponential relations of the form \( y = ae^{bx} \), taking the logarithmic derivative \( \frac{d}{dx}(\ln y) = b \) can often simplify the process of eliminating constants.

Question 4. For the D.E. of the family of curve \( y = Ae^{2x} + Be^{-3x} \).
Answer: We are given the relation: \[ y = Ae^{2x} + Be^{-3x} \quad \dots (i) \] Differentiating with respect to \( x \): \[ \frac{dy}{dx} = 2Ae^{2x} - 3Be^{-3x} \] Substituting \( Be^{-3x} = y - Ae^{2x} \) from equation (i): \[ \frac{dy}{dx} = 2Ae^{2x} - 3(y - Ae^{2x}) \]
\( \implies \frac{dy}{dx} = 5Ae^{2x} - 3y \quad \dots (ii) \) Differentiating again with respect to \( x \): \[ \frac{d^2y}{dx^2} = 10Ae^{2x} - 3 \frac{dy}{dx} \quad \dots (iii) \] Expressing \( Ae^{2x} \) from equation (ii) as \( Ae^{2x} = \frac{1}{5}\left(\frac{dy}{dx} + 3y\right) \) and substituting this into equation (iii): \[ \frac{d^2y}{dx^2} = 10 \left[ \frac{1}{5}\left(\frac{dy}{dx} + 3y\right) \right] - 3 \frac{dy}{dx} \]
\( \implies \frac{d^2y}{dx^2} = 2 \frac{dy}{dx} + 6y - 3 \frac{dy}{dx} \)
\( \implies \frac{d^2y}{dx^2} + \frac{dy}{dx} = 6y \) On simplifying, we get the required differential equation.
In simple words: We can get rid of the constants \( A \) and \( B \) by differentiating twice, expressing one exponential term in terms of the other variables, and substituting it back.

Exam Tip: For any equation of the form \( y = Ae^{mx} + Be^{nx} \), the resulting differential equation is always \( \frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0 \). Knowing this shortcut helps verify your final answer quickly!

Question 5. Form the D.E. of the family of curve \( y = e^x(A \cos x + B \sin x) \).
Answer: The equation of the family of curves is: \[ y = e^x(A \cos x + B \sin x) \quad \dots (i) \] Differentiating both sides with respect to \( x \) using the product rule: \[ \frac{dy}{dx} = e^x(-A \sin x + B \cos x) + e^x(A \cos x + B \sin x) \ ] Using equation (i), this can be written as: \[ \frac{dy}{dx} = e^x(-A \sin x + B \cos x) + y \quad \dots (ii) \] Differentiating again with respect to \( x \): \[ \frac{d^2y}{dx^2} = e^x(-A \cos x - B \sin x) + e^x(-A \sin x + B \cos x) + \frac{dy}{dx} \ ] Substituting the expressions from equations (i) and (ii): \[ \frac{d^2y}{dx^2} = -y + \left( \frac{dy}{dx} - y \right) + \frac{dy}{dx} \]
\( \implies \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \) This simplifies to the final differential equation.
In simple words: Use the product rule to differentiate. Substitute the original function \( y \) and its first derivative back into the second derivative to eliminate the arbitrary constants \( A \) and \( B \).

Exam Tip: When a family of curves involves trigonometric terms multiplied by an exponential, like \( e^{ax}(A\cos bx + B\sin bx) \), the corresponding differential equation is always \( y'' - 2ay' + (a^2+b^2)y = 0 \).

Question 6. Find the D.E. of all the circle touching x - axis at the origin.
Answer: The general equation of a circle that touches the x-axis at the origin is: \[ (x - 0)^2 + (y - k)^2 = k^2 \quad \dots (i) \] Here, \( k \) represents the variable parameter. Differentiating this equation with respect to \( x \): \[ 2x + 2(y - k)\frac{dy}{dx} = 0 \]
\( \implies y - k = -\frac{x}{\frac{dy}{dx}} \)
\( \implies k = y + \frac{x}{\frac{dy}{dx}} \) Substituting the expressions for \( y - k \) and \( k \) back into the circle's equation (i): \[ x^2 + \left( -\frac{x}{\frac{dy}{dx}} \right)^2 = \left( y + \frac{x}{\frac{dy}{dx}} \right)^2 \]
\( \implies x^2 + \frac{x^2}{\left( \frac{dy}{dx} \right)^2} = y^2 + \frac{x^2}{\left( \frac{dy}{dx} \right)^2} + \frac{2xy}{\frac{dy}{dx}} \) Cancelling the common term and simplifying gives the required differential equation: \[ x^2 - y^2 = \frac{2xy}{\frac{dy}{dx}} \]
\( \implies (x^2 - y^2)\frac{dy}{dx} = 2xy \)
In simple words: A circle touching the x-axis at the origin has its center on the y-axis, represented by \( (0, k) \). We set up its equation, take the derivative, and eliminate \( k \) to find the differential equation.

Exam Tip: Ensure you correctly identify the center of the circle. Since it touches the x-axis at the origin, its center must lie on the y-axis, making its coordinates \( (0, k) \) and its radius \( |k| \).

Question 7. Form the differential equation of family of circles in the second quadrant and touching the coordinate axis.
Answer: For a circle in the second quadrant that touches both coordinate axes, the equation is given by: \[ (x + a)^2 + (y - a)^2 = a^2 \quad \dots (i) \] Differentiating this equation with respect to \( x \): \[ 2(x + a) + 2(y - a)y' = 0 \]
\( \implies x + a + yy' - ay' = 0 \)
\( \implies x + yy' = a(y' - 1) \) Solving for the parameter \( a \), we get: \[ a = \frac{x + yy'}{y' - 1} \] Substituting this expression for \( a \) back into our initial circle equation (i): \[ \left( x + \frac{x + yy'}{y' - 1} \right)^2 + \left( y - \frac{x + yy'}{y' - 1} \right)^2 = \left( \frac{x + yy'}{y' - 1} \right)^2 \] Expanding and simplifying the terms inside the brackets: \[ \left( \frac{y'(x + y)}{y' - 1} \right)^2 + \left( \frac{-(x + y)}{y' - 1} \right)^2 = \left( \frac{x + yy'}{y' - 1} \right)^2 \] Since the denominator \( (y' - 1)^2 \) is common on both sides, we cancel it: \[ (x + y)^2 (y'^2 + 1) = (x + yy')^2 \]
\( \implies (x + y)^2 [1 + (y')^2] = (x + yy')^2 \) which represents the required differential equation.
In simple words: Circles in the second quadrant touching both axes have their center at \( (-a, a) \) and radius \( a \). Differentiate this equation to find \( a \) in terms of \( x, y, \) and \( y' \), then substitute it back to eliminate \( a \).

Exam Tip: Be careful with the signs when writing the equation of the circle in different quadrants. In the second quadrant, the x-coordinate of the center is negative (\( -a \)), while the y-coordinate is positive (\( a \)).

Question 8. Obtain the D.E. of all the circles with radius \( r \).
Answer: Let centre is \( (a, b) \) the equation of circle is
\( (x - a)^2 + (y - b)^2 = r^2 \quad \dots (i) \)
Diff. w.r.t. \( x \),
\( 2(x - a) + 2(y - b)\frac{dy}{dx} = 0 \)
\( \implies x - a + (y - b)\frac{dy}{dx} = 0 \quad \dots (ii) \)
Diff. again w.r.t. \( x \),
\( \implies 1 + (y - b)\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 \)
\( \implies x - a = \frac{\left[1+\left(\frac{dy}{dx}\right)^2\right] \cdot \frac{dy}{dx}}{\left(\frac{d^2y}{dx^2}\right)} \)
Put value of \( (x - a) \) & \( (y - b) \) in eq. (i) we get
\( \frac{\left(1+\left(\frac{dy}{dx}\right)^2\right)^2 \cdot \left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2y}{dx^2}\right)^2} + \frac{\left(1+\left(\frac{dy}{dx}\right)^2\right)^2}{\left(\frac{d^2y}{dx^2}\right)^2} = r^2 \)
\( \implies \left(1 + \left(\frac{dy}{dx}\right)^2\right)^2 \left(\left(\frac{dy}{dx}\right)^2 + 1\right) = r^2 \left(\frac{d^2y}{dx^2}\right)^2 \)
\( \implies \left(1 + \left(\frac{dy}{dx}\right)^2\right)^3 = r^2 \left(\frac{d^2y}{dx^2}\right)^2 \) is the required D.E.

Question 9. Find the D.E. of the family of parabolas having their axis of symmetry coincident with the axis of x.
Answer: The vertex is of the form \( (h, 0) \) since the axis of symmetry is the x-axis. The general equation for this family of parabolas is: \[ y^2 = 4a(x - h) \quad \dots (i) \] Here, \( a \) and \( h \) are the parameters to be eliminated. Differentiating both sides with respect to \( x \): \[ 2y \frac{dy}{dx} = 4a \]
\( \implies y \frac{dy}{dx} = 2a \quad \dots (ii) \) Differentiating once more to eliminate the constant \( a \): \[ y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0 \]
In simple words: Parabolas symmetric about the x-axis have their equation as \( y^2 = 4a(x - h) \). Taking the first derivative removes the vertex offset \( h \), and taking the second derivative removes the focus parameter \( a \).

Exam Tip: Note that there are two parameters, \( a \) and \( h \). Therefore, the differential equation must be of the second order, which we achieve by differentiating twice.

Question 10. Form the D.E. of the family of ellipses having foci on y - axis and centre at the origin.
Answer: The general equation for an ellipse with its center at the origin and foci on the y-axis is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \dots (i) \] Here, \( a \) and \( b \) are arbitrary constants to be eliminated. Differentiating this equation with respect to \( x \): \[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \]
\( \implies \frac{x}{a^2} + \frac{y}{b^2} \frac{dy}{dx} = 0 \quad \dots (ii) \) Differentiating a second time with respect to \( x \): \[ \frac{1}{a^2} + \frac{y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2}{b^2} = 0 \quad \dots (iii) \] Using equation (ii), we can express \( \frac{1}{a^2} = -\frac{y}{b^2 x} \frac{dy}{dx} \). Substituting this into equation (iii): \[ -\frac{y \frac{dy}{dx}}{b^2 x} + \frac{y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2}{b^2} = 0 \] Multiplying the entire equation by \( b^2 x \) and rearranging the terms gives the required differential equation: \[ -y \frac{dy}{dx} + x \left[ y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right] = 0 \]
\( \implies xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0 \)
In simple words: Differentiate the equation of the ellipse twice. Use the first derivative to substitute and eliminate the parameters \( a \) and \( b \) from the second derivative.

Exam Tip: Note that the resulting differential equation is identical for ellipses with foci on either the x-axis or the y-axis. The orientation does not change the form of the differential equation since both are eliminated during differentiation.

Please click the link below to download CBSE Class 12 Mathematics Linear Differential Equations (6).