CBSE Class 12 Mathematics Linear Differential Equations Worksheet Set 05

CBSE Class 12 Mathematics Linear Differential Equations (5). 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.

Question. Show that the general solution of the D.E. \( \frac{dy}{dx} + \frac{y^2+y+1}{x^2+x+1} = 0 \) is given by \( x + y + 1 = A(1 - x - y - 2xy) \) where \( A \) is the parameter.
Answer: We have, \( \frac{dy}{dx} = -\frac{y^2+y+1}{x^2+x+1} \)
\( \implies \frac{dy}{y^2+y+1} = \frac{-dx}{x^2+x+1} \)
Interpreting both sides:
\( \int \frac{dy}{y^2+y+1} = -\int \frac{dx}{x^2+x+1} \)
\( \implies \int \frac{1}{\left(y+\frac{1}{2}\right)^2 - \frac{1}{4} + 1} dy = -\int \frac{1}{\left(x+\frac{1}{2}\right)^2 - \frac{1}{4} + 1} dx \)
\( \implies \int \frac{1}{\left(y+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dy = -\int \frac{1}{\left(x+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dx \)
\( \implies \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y+1}{\sqrt{3}}\right) = -\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + c \)
\( \implies \frac{2}{\sqrt{3}}\left(\tan^{-1}\left(\frac{2y+1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)\right) = c \)
\( \implies \tan^{-1}\left( \frac{\frac{2y+1}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}}{1 - \left(\frac{2y+1}{\sqrt{3}}\right)\left(\frac{2x+1}{\sqrt{3}}\right)} \right) = \frac{\sqrt{3}}{2}c \)
\( \implies \tan^{-1}\left( \frac{\frac{2x+2y+2}{\sqrt{3}}}{\frac{3-4xy-2y-2x-1}{3}} \right) = \frac{\sqrt{3}}{2}c \)
\( \implies \frac{(2x+2y+2)\sqrt{3}}{2-4xy-2x-2y} = \tan \left( \frac{\sqrt{3}}{2}c \right) \)
\( \implies \frac{(x+y+1)\sqrt{3}}{1-2xy-x-y} = \tan \left( \frac{\sqrt{3}}{2}c \right) \)
\( \implies \frac{x+y+1}{1-2xy-x-y} = \frac{1}{\sqrt{3}}\tan\left(\frac{\sqrt{3}}{2}c\right) \)
\( \implies \frac{x+y+1}{1-2xy-x-y} = A \); where \( A = \frac{1}{\sqrt{3}}\tan\left(\frac{\sqrt{3}}{2}c\right) \)
\( \implies (x + y + 1) = A(1 - 2xy - x - y) \) is the required solution.

Question. Find the particular solution of the D.E. \( (1 + e^{2x})dy + (1 + y^2)e^xdx = 0 \); given \( x = 0, y = 1 \)
Answer: We have, \( (1 + e^{2x})dy = -(1 + y^2)e^xdx \)
\( \implies \frac{dy}{dx} = -\frac{(1+y^2)e^x}{1+e^{2x}} \)
Separating the variables & interpreting both sides:
\( \implies \int \frac{dy}{1+y^2} = -\int \frac{e^x}{1+e^{2x}} dx \)
Put \( e^x = t \implies e^x dx = dt \)
\( \implies \tan^{-1} y = -\int \frac{dt}{1+t^2} \)
\( \implies \tan^{-1} y = -\tan^{-1} t + c \)
\( \implies \tan^{-1}(y) + \tan^{-1}(e^x) = c \)
Put \( x = 0 \) & \( y = 1 \):
\( \implies \tan^{-1}(1) + \tan^{-1}(1) = c \implies c = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \)
\( \therefore \tan^{-1}(y) + \tan^{-1}(e^x) = \frac{\pi}{2} \)
\( \implies \tan^{-1}\left(\frac{y+e^x}{1-ye^x}\right) = \frac{\pi}{2} \)
\( \implies \frac{y+e^x}{1-ye^x} = \tan\left(\frac{\pi}{2}\right) \)
\( \implies \frac{y+e^x}{1-ye^x} = \frac{1}{0} \quad \dots \left\{\tan\left(\frac{\pi}{2}\right) = \infty\right\} \)
\( \implies 0 = 1 - ye^x \)
\( \implies y = \frac{1}{e^x} \) is the required solution.

Question. At any point \( (x, y) \) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point \( (-4, -3) \). Find the equation of the curve given that it passes through \( (-2,1) \).
Answer: It is given that \( (x, y) \) is the point of contact of the curve and its tangent.
Hence the slope of the line segment joining \( (x, y) \) and \( (-4, -3) \) is \( m_1 = \frac{y+3}{x+4} \).
Let this be \( m_1 \)
But we know that the slope of the tangent to the curve is \( \frac{dy}{dx} \).
Let this be \( m_2 \)
According to the given information, \( m_2 = 2m_1 \)
\( \frac{dy}{dx} = \frac{2(y+3)}{x+4} \)
Separating the variables, we get:
\( \frac{dy}{y+3} = \frac{2dx}{x+4} \)
Integrating on both sides, we get:
\( \int \frac{dy}{y+3} = 2\int \frac{dx}{x+4} \)
\( \implies \log(y + 3) = 2 \log(x + 4) + \log c \)
\( \implies \log(y + 3) = \log [c(x + 4)^2] \)
\( \implies y + 3 = c(x + 4)^2 \)
It is given that the curve passes through the point \( (-2, 1) \).
Substituting for \( x \) and \( y \) in the general equation to evaluate for \( c \), we get:
\( 1 + 3 = c(-2 + 4)^2 \)
\( \implies 4 = 4c \)
\( \implies c = 1 \)
Substituting this for \( c \), we get:
\( y + 3 = (x + 4)^2 \) is the required equation of the curve.

Question. Solve the D.E. \( \sqrt{1 + x^2 + y^2 + x^2y^2} + xy\frac{dy}{dx} = 0 \)
Answer: We have, \( xy\frac{dy}{dx} = -\sqrt{1 + x^2 + y^2 + x^2y^2} \)
\( \implies \frac{dy}{dx} = -\frac{\sqrt{(1+x^2)(1+y^2)}}{xy} \)
\( \implies \frac{dy}{dx} = -\frac{\sqrt{1+x^2}\sqrt{1+y^2}}{xy} \)
Separating the variables & interpreting both sides:
\( \implies \int \frac{y}{\sqrt{1+y^2}} dy = -\int \frac{\sqrt{1+x^2}}{x} dx \)
Put \( 1+y^2 = t \implies ydy = \frac{dt}{2} \)
Put \( 1+x^2 = z^2 \implies 2xdx = 2zdz \implies dx = \frac{zdz}{x} \)
\( \therefore \frac{1}{2}\int \frac{dt}{\sqrt{t}} = -\int \frac{z}{x} \cdot \frac{zdx}{x} \)
\( \implies \frac{1}{2} \times 2\sqrt{t} = -\int \frac{z^2}{z^2-1} dz \)
\( \implies \sqrt{t} = -\int \frac{z^2-1+1}{z^2-1} dz \)
\( \implies \sqrt{t} = -\int \left( 1 + \frac{1}{z^2-1} \right) dz \)
\( \implies \sqrt{1+y^2} = -\left[ z + \frac{1}{2} \log \left| \frac{z-1}{z+1} \right| \right] + c \)
\( \implies \sqrt{1+y^2} = -\left[ \sqrt{1+x^2} + \frac{1}{2} \log \left| \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1} \right| \right] + c \)
\( \implies \sqrt{1+x^2} + \sqrt{1+y^2} + \frac{1}{2} \log \left| \frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1} \right| = c \)

Question. Find the equation of the curve passing through the point \( (1,0) \) given that slope of the tangent to the curve at any point \( (x, y) \) is \( \frac{2x(\log x+1)}{\sin y+y \cos y} \).
Answer: Slope of tangent at any point \( (x, y) \) is given by \( \frac{dy}{dx} \).
We have, \( \frac{dy}{dx} = \frac{2x(\log x+1)}{\sin y+y \cos y} \)
\( \implies (\sin y + y \cos y)dy = 2x(\log x + 1)dx \)
Interpreting both sides:
\( \int(\sin y + y \cos y)dy = 2 \int x(\log x + 1)dx \)
\( \implies \int \sin y dy + \int y \cos y dy = 2 \int x dx + 2 \int x \log x dx \)
\( \implies -\cos y + y \sin y - \int \sin y dy = \frac{2x^2}{2} + 2 \left[ \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx \right] \)
\( \implies -\cos y + y \sin y + \cos y = x^2 + 2 \left( \frac{x^2}{2} \log x - \frac{x^2}{4} \right) + c \)
\( \implies y \sin y = x^2 + x^2 \log x - \frac{x^2}{2} + c \)
This equation passes through the point \( (1,0) \).
Put \( x = 1 \) and \( y = 0 \):
\( \implies 0 = 1 + 0 - \frac{1}{2} + c \)
\( \implies c = -\frac{1}{2} \)
\( \therefore y \sin y = x^2 + x^2 \log x - \frac{x^2}{2} - \frac{1}{2} \)
\( \implies y \sin y = \frac{x^2}{2} + x^2 \log x - \frac{1}{2} \)
\( \implies 2y \sin y = x^2 + 2x^2 \log x - 1 \) is the required equation of curve.

Question. Solve the D.E. \( 3e^x \tan y dx + (1 - e^x) \sec^2 y dy = 0 \)
Answer: We have, \( (1 - e^x) \sec^2 y dy = -3e^x \tan y dx \)
\( \implies \frac{dy}{dx} = \frac{-3e^x \tan y}{(1-e^x) \sec^2 y} \)
Separating variables & interpreting both sides:
\( \implies \int \frac{\sec^2 y}{\tan y} dy = -3 \int \frac{e^x}{1-e^x} dx \)
Put \( \tan y = t \implies \sec^2 y dy = dt \)
Put \( 1 - e^x = z \implies -e^x dx = dz \)
\( \implies \log|t| = 3 \log z + \log c \)
\( \implies \log\left| \frac{t}{z^3} \right| = \log c \)
Replacing \( t \) & \( z \):
\( \implies \left| \frac{\tan y}{(e^x-1)^3} \right| = c \)
\( \implies \frac{\tan y}{(e^x-1)^3} = \pm c \)
\( \implies \tan y = c_1(e^x - 1)^3 \) where \( c_1 = \pm c \) is the required solution.

Question. For the D.E. \( xy\frac{dy}{dx} = (x + 2)(y + 2) \). Find the solution curve passes through \( (1,-1) \).
Answer: We have, \( \frac{dy}{dx} = \frac{(x+2)(y+2)}{xy} \)
\( \implies \frac{y}{y+2} dy = \frac{x+2}{x} dx \)
Interpreting both sides:
\( \implies \int \frac{y+2-2}{y+2} dy = \int \left( \frac{x}{x} + \frac{2}{x} \right) dx \)
\( \implies \int \left( 1 - \frac{2}{y+2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx \)
\( \implies y - 2 \log|y + 2| = x + 2 \log|x| + c \)
It passes through the point \( (1,-1) \).
Put \( x = 1 \) & \( y = -1 \):
\( \therefore -1 - 2 \log|-1+2| = 1 + 2 \log|1| + c \)
\( \implies -1 - 0 = 1 + 0 + c \implies c = -2 \)
\( \therefore y - 2 \log|y + 2| = x + 2 \log|x| - 2 \)
\( \implies y - x + 2 = 2 \log|x| + 2 \log|y+2| \)
\( \implies y - x + 2 = \log|x^2(y + 2)^2| \)
\( \implies y - x + 2 = \log\left(x^2(y + 2)^2\right) \)

Question. In a bank principal increases at the rate of 5% per year. In how many years Rs.1000 double itself.
Answer: Let \( P \) be the principal at any time \( t \).
Then, according to the question:
\( \frac{dp}{dt} = 5 \frac{p}{100} \)
\( \implies \frac{dp}{dt} = \frac{p}{20} \)
\( \implies \frac{1}{p} dp = \frac{1}{20} dt \)
Interpreting both sides:
\( \int \frac{1}{p} dp = \frac{1}{20} \int dt \)
\( \implies \log p = \frac{1}{20}t + c \)
\( \implies p = e^{\frac{1}{20}t+c} \)
\( \implies p = e^{\frac{t}{20}} \cdot e^c \)
\( \implies p = e^{\frac{t}{20}} \cdot c_1 \) where \( c_1 = e^c \)
Given at \( t = 0 \); \( p = 1000 \):
\( \therefore 1000 = e^0 \cdot c_1 \)
\( \implies c_1 = 1000 \)
\( \therefore p = e^{\frac{t}{20}} \cdot 1000 \)
Let at \( t = t_1 \); \( p = 2000 \):
\( \implies 2000 = e^{\frac{t_1}{20}} \cdot 1000 \)
\( \implies e^{\frac{t_1}{20}} = 2 \)
Taking log on both sides:
\( \frac{t_1}{20} = \log_e 2 \)
\( t_1 = 20 \log_e 2 \) years
\( \therefore \) principal doubles in \( 20 \log_e 2 \) years.

Question. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units & after 3 seconds it is 6 units. Find the radius of the balloon after \( t \) seconds.
Answer: Let the rate of change of the volume of balloon be \( k \).
Hence, \( \frac{dv}{dt} = k \)
\( \implies \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = k \)
\( \implies \frac{4}{3}\pi(3r^2) \left(\frac{dr}{dt}\right) = k \)
Now separating the variables, we get:
\( 4\pi r^2 dr = k dt \)
Integrating on both sides we get:
\( 4\pi \int r^2 dr = k \int dt \)
\( \implies 4\pi \left(\frac{r^3}{3}\right) = kt + C \)
\( \implies 4\pi r^3 = 3(kt + C) \)
Given \( t = 0, r = 3 \):
\( 4\pi (3^3) = 3(k \cdot 0 + C) \)
\( \implies 108\pi = 3C \)
\( \implies C = 36\pi \)
When \( t = 3, r = 6 \):
\( 4\pi \cdot 6^3 = 3(3k + C) \)
\( \implies 864\pi = 3(3k + 36\pi) \)
Dividing throughout by 3 we get:
\( 3k = 288\pi - 36\pi \)
\( \implies 3k = 252\pi \)
Hence, \( k = 84\pi \).
Now substituting the values of \( k \) and \( C \):
\( 4\pi r^3 = 3(84\pi t + 36\pi) \)
Taking \( 12\pi \) as a common factor:
\( 4\pi r^3 = 4\pi (63t + 27) \)
Dividing throughout by \( 4\pi \):
\( r^3 = 63t + 27 \)
\( r = (63t + 27)^{\frac{1}{3}} \)
Thus the radius of the balloon after \( t \) seconds is \( (63t + 27)^{\frac{1}{3}} \).

Question. In a culture the bacteria count is 1,00,000. The number is increases by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Answer: From the given information we know that \( \frac{dy}{dt} \) is proportional to \( y \).
\( \frac{dy}{dt} = ky \)
On separating the variables, we get:
\( \frac{dy}{y} = k \cdot dt \)
Integrating on both sides, we get:
\( \int \frac{dy}{y} = k \int dt \)
\( \log y = kt + c \quad \dots (1) \)
Let \( y_0 \) be the number of bacteria when \( t = 0 \).
Hence, \( \log y_0 = c \)
Substituting this value in equation (1), we get:
\( \log y = kt + \log y_0 \)
\( \implies \log y - \log y_0 = kt \)
\( \implies \log \left(\frac{y}{y_0}\right) = kt \quad \dots (2) \)
It is also given that the number of bacteria increases 10% in 2 hours.
Hence, \( y = \frac{110}{100} y_0 = \frac{11}{10} y_0 \) at \( t = 2 \).
Substituting this in (2):
we get \( 2k = \log\left(\frac{11}{10}\right) \)
or \( k = \frac{1}{2} \log\left(\frac{11}{10}\right) \)
Therefore, \( \frac{1}{2} \log\left(\frac{11}{10}\right) t = \log\left(\frac{y}{y_0}\right) \)
\( \implies t = \frac{2 \log\left(\frac{y}{y_0}\right)}{\log\left(\frac{11}{10}\right)} \)
Let the time when the number of bacteria increases from 1,00,000 to 2,00,000 be \( t_1 \).
\( y = 2y_0 \) at \( t = t_1 \)
Hence, \( t_1 = \frac{2 \log\left(\frac{2y_0}{y_0}\right)}{\log\left(\frac{11}{10}\right)} = \frac{2 \log 2}{\log\left(\frac{11}{10}\right)} \)
Hence, in \( \frac{2 \log 2}{\log\left(\frac{11}{10}\right)} \) hours, the number of bacteria increases from 1,00,000 to 2,00,000.

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