CBSE Class 12 Mathematics Linear Differential Equations Worksheet Set 04

CBSE Class 12 Mathematics Linear Differential Equations (4). 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. Solve the D.E. \( ydx + x \log y dy - x \log x dy - 2x dy = 0 \)
Answer: We have, \( y dx + x dy(\log y - \log x) - 2x dy = 0 \)
\( \implies y dx + x dy \cdot \log\left(\frac{y}{x}\right) - 2x dy = 0 \)
\( \implies dy\left(x \log\left(\frac{y}{x}\right) - 2x\right) = -y dx \)
\( \implies \frac{dy}{dx} = -\frac{y}{x \log\left(\frac{y}{x}\right) - 2x} \quad \dots (i) \)
It is a homogeneous D.E.
Put \( y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx} \), substituting in eq. (i):
\( v + x\frac{dv}{dx} = -\frac{vx}{x \log v - 2x} \)
\( \implies v + x\frac{dv}{dx} = -\frac{v}{\log v - 2} \)
\( \implies x\frac{dv}{dx} = -\frac{v}{\log v - 2} - v \)
\( \implies x\frac{dv}{dx} = \frac{-v - v \log v + 2v}{\log v - 2} \)
\( \implies x\frac{dv}{dx} = \frac{v - v \log v}{\log v - 2} \)
\( \implies x\frac{dv}{dx} = -\frac{v(\log v - 1)}{\log v - 2} \)
\( \implies \frac{\log v - 2}{v(\log v - 1)} dv = -\frac{dx}{x} \)
Interpreting both sides:
\( \int \frac{\log v - 2}{v(\log v - 1)} dv = -\int \frac{dx}{x} \)
Put \( \log v - 1 = t \implies \frac{1}{v} dv = dt \)
\( \therefore \int \frac{t - 1}{t} dt = -\log|x| \)
\( \implies \int \left(1 - \frac{1}{t}\right) dt = -\log|x| \)
\( \implies t - \log t = -\log x + c \)
\( \implies \log v - 1 - \log(\log v - 1) = -\log x + c \)
\( \implies \log\left(\frac{vx}{\log v - 1}\right) = c + 1 \)
Replace \( v \) by \( \frac{y}{x} \):
\( \implies \log\left(\frac{y}{\log\left(\frac{y}{x}\right) - 1}\right) = c + 1 \)
\( \implies \frac{y}{\log\left(\frac{y}{x}\right) - 1} = e^{c+1} \)
\( \implies \frac{y}{\log\left(\frac{y}{x}\right) - 1} = \pm e^{c+1} \)
\( \implies y = c_1 \left(\log\left(\frac{y}{x}\right) - 1\right) \); where \( c_1 = \pm e^{c+1} \) is the required general solution.

Question. Solve the D.E. \( \left(xe^{\frac{y}{x}} + y\right) dx = xdy; y(1) = 1 \)
Answer: We have, \( \frac{dy}{dx} = \frac{xe^{\frac{y}{x}} + y}{x} \quad \dots (i) \)
Note: it is a homogeneous D.E.
Put \( y = vx \)
Differentiating w.r.t. \( x \), \( \frac{dy}{dx} = v + x\frac{dv}{dx} \), substituting in eq. (i):
\( v + x\frac{dv}{dx} = \frac{xe^v + vx}{x} \)
\( \implies v + x\frac{dv}{dx} = e^v + v \)
\( \implies x\frac{dv}{dx} = e^v \)
\( \implies e^{-v} dv = \frac{dx}{x} \)
\( \implies \int e^{-v} dv = \int \frac{dx}{x} \)
\( \implies -e^{-v} = \log|x| + c \)
\( \implies e^{-v} + \log|x| = -c \)
Replace \( v \) by \( \frac{y}{x} \):
\( e^{-\frac{y}{x}} + \log|x| = -c \)
Put \( x = 1 \) and \( y = 1 \):
\( \implies e^{-1} + \log|1| = -c \)
\( \implies \frac{1}{e} = -c \), substituting this in the above equation:
\( \implies e^{-\frac{y}{x}} + \log|x| = \frac{1}{e} \) is the required solution.

Question. Solve the D.E. \( xy \log\left(\frac{x}{y}\right) dx + \left(y^2 - x^2 \log\left(\frac{x}{y}\right)\right) dy = 0 \)
Answer: We have, \( xy \log\left(\frac{x}{y}\right) dx = -\left(y^2 - x^2 \log\left(\frac{x}{y}\right)\right) dy \)
\( \implies \frac{dx}{dy} = \frac{-\left(y^2 - x^2 \log\left(\frac{x}{y}\right)\right)}{xy \log\left(\frac{x}{y}\right)} \)
It is a homogeneous D.E.
Put \( x = vy \)
Differentiating w.r.t. \( y \), \( \frac{dx}{dy} = v + y\frac{dv}{dy} \):
\( v + y\frac{dv}{dy} = \frac{-\left(y^2 - v^2 y^2 \log v\right)}{vy^2 \log v} \)
\( \implies v + y\frac{dv}{dy} = \frac{-\left(1 - v^2 \log v\right)}{v \log v} \)
\( \implies y\frac{dv}{dy} = \frac{-\left(1 - v^2 \log v\right)}{v \log v} - v \)
\( \implies y\frac{dv}{dy} = \frac{-1 + v^2 \log v - v^2 \log v}{v \log v} \)
\( \implies y\frac{dv}{dy} = \frac{-1}{v \log v} \)
\( \implies v \log v dv = -\frac{dy}{y} \)
\( \implies \int v \log v dv = -\int \frac{dy}{y} \)
\( \implies \log v \cdot \frac{v^2}{2} - \int \frac{1}{v} \cdot \frac{v^2}{2} dv = -\log|y| \)
\( \implies \frac{v^2}{2} \log v - \frac{1}{2} \int v dv = -\log|y| \)
\( \implies \frac{v^2}{2} \log v - \frac{v^2}{4} = -\log|y| + c \)
\( \implies v^2 \log v - \frac{v^2}{2} = -2\log|y| + 2c \)
Replace \( v \) by \( \frac{x}{y} \):
\( \implies \frac{x^2}{y^2}\log\left(\frac{x}{y}\right) - \frac{1}{2}\left(\frac{x^2}{y^2}\right) = -2\log|y| + 2c \)
\( \implies \frac{x^2}{y^2}\left(\log\left(\frac{x}{y}\right) - \frac{1}{2}\right) + \log|y^2| = 2c \) is the required solution.

Question. Solve the D.E. \( \sin^{-1}\left(\frac{dy}{dx}\right) = x + y \)
Answer: We have, \( \sin^{-1}\left(\frac{dy}{dx}\right) = x + y \)
\( \implies \frac{dy}{dx} = \sin(x + y) \)
Put \( x + y = v \)
Differentiating w.r.t. \( x \), \( 1 + \frac{dy}{dx} = \frac{dv}{dx} \)
\( \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
\( \therefore \frac{dv}{dx} - 1 = \sin v \)
\( \implies \frac{dv}{dx} = 1 + \sin v \)
\( \implies \int \frac{dv}{1+\sin v} = \int dx \)
\( \implies \int \frac{1}{1+\sin v} \times \frac{1-\sin v}{1-\sin v} dv = \int dx \) (Rationalizing)
\( \implies \int \frac{1-\sin v}{\cos^2 v} dv = x \)
\( \implies \int \left(\sec^2 v - \tan v \cdot \sec v\right) dv = x \)
\( \implies \tan v - \sec v = x + c \)
Replace \( v \) by \( x + y \):
\( \therefore \tan(x + y) - \sec(x + y) = x + c \) is the required solution.

Question. Solve the D.E. \( (x + y)^2 \frac{dy}{dx} = a^2 \)
Answer: We have, \( \frac{dy}{dx} = \frac{a^2}{(x+y)^2} \quad \dots (i) \)
Put \( x + y = v \)
Differentiating w.r.t. \( x \), \( 1 + \frac{dy}{dx} = \frac{dv}{dx} \)
\( \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
\( \therefore \) equation (i) becomes:
\( \frac{dv}{dx} - 1 = \frac{a^2}{v^2} \)
\( \implies \frac{dv}{dx} = \frac{a^2}{v^2} + 1 \)
\( \implies \frac{dv}{dx} = \frac{a^2+v^2}{v^2} \)
\( \implies \frac{v^2}{v^2+a^2} dv = dx \)
Interpreting both sides:
\( \int \frac{v^2}{v^2+a^2} dv = \int dx \)
\( \implies \int \frac{v^2+a^2-a^2}{v^2+a^2} dv = \int dx \)
\( \implies \int \left(1 - \frac{a^2}{v^2+a^2}\right) dv = \int dx \)
\( \implies v - a^2 \times \frac{1}{a} \tan^{-1}\left(\frac{v}{a}\right) = x + c \)
Replace \( v \) by \( x + y \):
\( \implies x + y - a \tan^{-1}\left(\frac{x+y}{a}\right) = x + c \)
\( \implies y = a \tan^{-1}\left(\frac{x+y}{a}\right) + c \) is the required solution.

Question. Solve \( \frac{dy}{dx} = \cos(x + y) + \sin(x + y) \)
Answer: Put \( x + y = v \)
Differentiating w.r.t. \( x \), \( 1 + \frac{dy}{dx} = \frac{dv}{dx} \)
\( \implies \frac{dy}{dx} = \frac{dv}{dx} - 1 \)
\( \therefore \frac{dv}{dx} - 1 = \cos v + \sin v \)
\( \implies \frac{dv}{dx} = 1 + \cos v + \sin v \)
\( \implies \int \frac{1}{1+\sin v+\cos v} dv = \int dx \) (Type single \( \sin x \), \( \cos x \))
\( \implies \int \frac{1}{1 + \frac{2\tan\frac{v}{2}}{1+\tan^2\frac{v}{2}} + \frac{1-\tan^2\frac{v}{2}}{1+\tan^2\frac{v}{2}}} dv = x \)
\( \implies \int \frac{1+\tan^2\frac{v}{2}}{1+\tan^2\left(\frac{v}{2}\right) + 2\tan\left(\frac{v}{2}\right) + 1 - \tan^2\left(\frac{v}{2}\right)} dv = x \)
\( \implies \int \frac{\sec^2\left(\frac{v}{2}\right)}{2\left(1+\tan\frac{v}{2}\right)} dv = x \)
Put \( 1 + \tan\frac{v}{2} = t \)
\( \implies \sec^2\left(\frac{v}{2}\right) \cdot \frac{1}{2} dv = dt \)
\( \implies \int \frac{dt}{t} = x \)
\( \implies \log\left|1 + \tan\frac{v}{2}\right| = x + c \)
\( \implies \log\left|1 + \tan\left(\frac{x+y}{2}\right)\right| = x + c \) is the required solution.

Question. Solve the initial problem \( (x - y)(dx + dy) = dx - dy ; y(0) = -1 \)
Answer: We have, \( xdx + xdy - ydx - ydy = dx - dy \)
\( \implies dy(x - y + 1) = dx(1 - x + y) \)
\( \implies \frac{dy}{dx} = \frac{1-(x-y)}{x-y+1} \)
Put \( x - y = v \)
Differentiating w.r.t. \( x \), \( 1 - \frac{dy}{dx} = \frac{dv}{dx} \)
\( \implies \frac{dy}{dx} = 1 - \frac{dv}{dx} \)
\( \therefore 1 - \frac{dv}{dx} = \frac{1-v}{v+1} \)
\( \implies \frac{dv}{dx} = 1 - \frac{1-v}{v+1} \)
\( \implies \frac{dv}{dx} = \frac{v+1-1+v}{v+1} \)
\( \implies \frac{dv}{dx} = \frac{2v}{v+1} \)
\( \implies \frac{v+1}{v} dv = 2dx \)
Interpreting both sides:
\( \int \frac{v+1}{v} dv = 2\int dx \)
\( \implies \int \left(1 + \frac{1}{v}\right) dv = 2x \)
\( \implies v + \log|v| = 2x + c \)
\( \implies x - y + \log|x - y| = 2x + c \)
\( \implies \log|x - y| = x + y + c \)
Put \( x = 0 \) & \( y = -1 \):
\( \implies \log|1| = 0 - 1 + c \)
\( \implies 0 = -1 + c \implies c = 1 \)
\( \therefore \log|x - y| = x + y + 1 \)
\( \implies |x - y| = e^{x+y+1} \) is the required solution.

Question. Solve the D.E. \( (x + 1)\frac{dy}{dx} = 2e^{-y} - 1; y(0) = 0 \)
Answer: We have, \( \frac{dy}{dx} = \frac{2e^{-y}-1}{x+1} \)
Separating the variables:
\( \implies \frac{dy}{2e^{-y}-1} = \frac{dx}{x+1} \)
Interpreting both sides:
\( \int \frac{dy}{2e^{-y}-1} = \int \frac{dx}{x+1} \)
\( \implies \int \frac{e^y}{2-e^y} dy = \int \frac{dx}{x+1} \)
Put \( 2 - e^y = t \)
\( \implies -e^y dy = dt \)
\( \implies e^y dy = -dt \)
\( \implies -\log|2 - e^y| = \log|x + 1| + \log c \)
\( \implies \log\left|\frac{1}{2-e^y}\right| = \log|c(x + 1)| \)
\( \implies \left|\frac{1}{2-e^y}\right| = |c(x+1)| \)
Put \( x = 0 \) & \( y = 0 \):
\( \implies \left|\frac{1}{2-1}\right| = |c(0 + 1)| \implies c = 1 \)
\( \implies \left|\frac{1}{2-e^y}\right| = |x + 1| \)
\( \implies |(x + 1)(2 - e^y)| = 1 \)
\( \implies (x + 1)(2 - e^y) = \pm 1 \)
But \( x = 0, y = 0 \) does not satisfy \( (x+1)(2-e^y) = -1 \).
\( \therefore (x+1)(2-e^y) = 1 \)
\( \implies 2 - e^y = \frac{1}{x+1} \)
\( \implies e^y = 2 - \frac{1}{x+1} \)
\( \implies e^y = \frac{2x+1}{x+1} \)
\( \implies y = \log\left(\frac{2x+1}{x+1}\right) \) is the required solution.

Question. Solve the D.E. \( y - x\frac{dy}{dx} = a\left(y^2 + \frac{dy}{dx}\right) \)
Answer: We have, \( y - x\frac{dy}{dx} = ay^2 + a\frac{dy}{dx} \)
\( \implies \frac{dy}{dx}(a + x) = y - ay^2 \)
\( \implies \frac{dy}{dx} = \frac{-ay^2+y}{a+x} \)
Separating variables & interpreting both sides:
\( \int \frac{dy}{ay^2-y} = -\int \frac{dx}{a+x} \)
\( \implies \frac{1}{a} \int \frac{1}{y^2-\frac{y}{a}} dy = -\log|x + a| \)
\( \implies \frac{1}{a} \int \frac{1}{\left(y-\frac{1}{2a}\right)^2-\left(\frac{1}{2a}\right)^2} dy = -\log|x + a| \)
\( \implies \frac{1}{a} \times \frac{1}{2 \times \frac{1}{2a}} \log\left| \frac{y-\frac{1}{2a}-\frac{1}{2a}}{y-\frac{1}{2a}+\frac{1}{2a}} \right| = -\log|x + a| + \log c \)
\( \implies \log\left| \frac{ay-1}{ay} \right| + \log|x + a| = \log c \)
\( \implies \log\left| \frac{(ay-1)(x+a)}{ay} \right| = \log c \)
\( \implies \left| \frac{(ay-1)(x+a)}{ay} \right| = c \)
\( \implies \frac{(ay-1)(x+a)}{ay} = \pm c \)
\( \implies (ay - 1)(x + a) = \pm a c y \)
\( \implies (ay - 1)(x + a) = c_1 y \); where \( c_1 = \pm a c \) is the required solution.

Question. Solve the initial value problem \( \log\left(\frac{dy}{dx}\right) = 3x + 4y ; x = 0 \) and \( y = 0 \).
Answer: We have, \( \log\left(\frac{dy}{dx}\right) = 3x + 4y \)
\( \implies \frac{dy}{dx} = e^{3x+4y} \)
\( \implies \frac{dy}{dx} = e^{3x} \cdot e^{4y} \)
\( \implies \int e^{-4y} dy = \int e^{3x} dx \)
\( \implies -\frac{1}{4}e^{-4y} = \frac{1}{3}e^{3x} + c \)
Put \( x = 0 \) & \( y = 0 \):
\( \implies -\frac{1}{4} = \frac{1}{3} + c \)
\( c = -\frac{7}{12} \)
\( \therefore -\frac{1}{4}e^{-4y} = \frac{1}{3}e^{3x} - \frac{7}{12} \)
\( \implies -\frac{1}{4}e^{-4y} = \frac{4e^{3x}-7}{12} \)
\( \implies -3e^{-4y} = 4e^{3x} - 7 \)
\( \implies 4e^{3x} + 3e^{-4y} = 7 \) is the required solution.

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