CBSE Class 12 Mathematics Linear Differential Equations Worksheet Set 03

CBSE Class 12 Mathematics Linear Differential Equations (3). 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 DE \( (x + y)\frac{dy}{dx} = 1 \).
Answer: \( \frac{dy}{dx} = \frac{1}{x+y} \)
\( \implies \frac{dx}{dy} = x + y \)
\( \implies \frac{dx}{dy} - x = y \)
Comparing with \( \frac{dx}{dy} + Px = Q \)
Here \( P = -1 \) & \( Q = y \)
\( I.F. = e^{-\int 1 dy} = e^{-y} \)
Solution is given by:
\( x \cdot (I.F.) = \int Q \cdot (I.F.) dy + c \)
\( \implies x \cdot e^{-y} = \int y \cdot e^{-y} dy + c \)
\( \implies x \cdot e^{-y} = y \frac{e^{-y}}{-1} - \int 1 \cdot \frac{e^{-y}}{-1} dy + c \)
\( \implies x \cdot e^{-y} = -ye^{-y} + \int e^{-y} dy + c \)
\( \implies x \cdot e^{-y} = -ye^{-y} - e^{-y} + c \)
\( \implies x e^{-y} = -e^{-y}(y + 1) + c \)
\( \implies x = -(y + 1) + ce^y \)

Question. Solve the D.E. \( x\frac{dy}{dx} - y = (x + 1)e^{-x}; y(1) = 0 \).
Answer: Divide by \( x \):
\( \frac{dy}{dx} - \frac{y}{x} = \frac{x+1}{x}e^{-x} \)
Comparing with \( \frac{dy}{dx} + Py = Q \)
Here \( P = -\frac{1}{x} \) and \( Q = \frac{x+1}{x} \cdot e^{-x} \)
\( I.F. = e^{-\int \frac{1}{x} dx} = e^{-\log x} = e^{\log x^{-1}} = \frac{1}{x} \)
Solution is given by:
\( y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c \)
\( \implies y \cdot \frac{1}{x} = \int \frac{x+1}{x}e^{-x} \cdot \frac{1}{x} dx + c \)
\( \implies \frac{y}{x} = \int \left( \frac{1}{x} + \frac{1}{x^2} \right) e^{-x} dx + c \)
\( \implies \frac{y}{x} = \int e^{-x} \cdot \frac{1}{x} dx + \int e^{-x} \cdot \frac{1}{x^2} dx + c \)
\( \implies \frac{y}{x} = \frac{1}{x} \cdot \frac{e^{-x}}{(-1)} - \int \left(-\frac{1}{x^2}\right) \cdot \frac{e^{-x}}{(-1)} dx + \int e^{-x} \cdot \frac{1}{x^2} dx + c \)
\( \implies \frac{y}{x} = -\frac{1}{x}e^{-x} - \int \frac{1}{x^2} e^{-x} dx + \int \frac{1}{x^2} e^{-x} dx + c \)
\( \implies \frac{y}{x} = -\frac{1}{x}e^{-x} + c \)
Put \( x = 1 \) and \( y = 0 \):
\( \implies 0 = -e^{-1} + c \implies c = \frac{1}{e} \)
\( \therefore \frac{y}{x} = -\frac{1}{x}e^{-x} + \frac{1}{e} \)
\( \implies y = -e^{-x} + xe^{-1} \) is the required solution.

Question. Show that the D.E. is homogeneous D.E. & also find the particular solution. \( (x + y)dy + (x - y)dx = 0; y = 1 \) when \( x = 1 \)
Answer: We have, \( (x + y)dy + (x - y)dx = 0 \)
\( \implies \frac{dy}{dx} = \frac{-(x-y)}{x+y} = \frac{y-x}{x+y} \quad \dots (i) \)
Here \( f(x, y) = \frac{y-x}{x+y} \)
\( f(\lambda x, \lambda y) = \frac{\lambda y - \lambda x}{\lambda x + \lambda y} \)
\( f(\lambda x, \lambda y) = \frac{\lambda(y-x)}{\lambda(x+y)} \)
\( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \)
Clearly function is homogeneous function of degree 0.
\( \therefore \) D.E. is a homogeneous D.E.
Put \( y = vx \)
Differentiating w.r.t. \( x \), \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
\( \therefore \) equation (i) becomes:
\( v + x\frac{dv}{dx} = \frac{vx-x}{x+vx} \)
\( \implies v + x\frac{dv}{dx} = \frac{v-1}{1+v} \)
\( \implies x\frac{dv}{dx} = \frac{v-1}{1+v} - v \)
\( \implies x\frac{dv}{dx} = \frac{v-1-v-v^2}{1+v} \)
\( \implies x\frac{dv}{dx} = \frac{-(v^2+1)}{1+v} \)
\( \implies \frac{1+v}{v^2+1} dv = -\frac{dx}{x} \) (separating variables)
Interpreting both sides:
\( \implies \int \frac{1+v}{1+v^2} dv = -\int \frac{dx}{x} \)
Separating:
\( \implies \int \frac{1}{1+v^2} dv + \int \frac{v}{1+v^2} dv = -\log|x| \)
Put \( 1 + v^2 = t \implies vdv = \frac{dt}{2} \):
\( \implies \tan^{-1} v + \frac{1}{2} \int \frac{dt}{t} = -\log|x| \)
\( \implies \tan^{-1} v + \frac{1}{2}\log|v^2 + 1| = -\log|x| + c \)
\( \implies 2\tan^{-1} v + \log|v^2 + 1| = -2\log|x| + 2c \)
\( \implies 2\tan^{-1} v + \log|v^2 + 1| + \log|x|^2 = 2c \)
Replace \( v \) by \( \frac{y}{x} \):
\( \implies 2\tan^{-1}\left(\frac{y}{x}\right) + \log\left| \left( \frac{y^2}{x^2} + 1 \right) \cdot x^2 \right| = 2c \)
\( \implies 2\tan^{-1}\left(\frac{y}{x}\right) + \log|x^2 + y^2| = 2c \)
Put \( x = 1 \) and \( y = 1 \):
\( \implies 2\tan^{-1}(1) + \log(2) = 2c \)
\( \implies 2\left(\frac{\pi}{4}\right) + \log(2) = 2c \)
\( \implies 2c = \frac{\pi}{2} + \log 2 \)
Solution is given by:
\( \therefore 2\tan^{-1}\left(\frac{y}{x}\right) + \log|x^2 + y^2| = \frac{\pi}{2} + \log 2 \).

Question. Show that D.E. is homogeneous & also find the initial value problem \( (x^2 + xy)dy = (x^2 + y^2)dx \) given \( y(1) = 0 \).
Answer: We have, \( \frac{dy}{dx} = \frac{x^2+y^2}{x^2+xy} \quad \dots (i) \)
Here \( f(x, y) = \frac{x^2+y^2}{x^2+xy} \)
\( f(\lambda x, \lambda y) = \frac{\lambda^2 x^2 + \lambda^2 y^2}{\lambda^2 x^2 + \lambda^2 xy} \)
\( f(\lambda x, \lambda y) = \lambda^0 \cdot f(x, y) \)
Clearly function is homogeneous of degree 0.
\( \therefore \) D.E. is homogeneous D.E.
Now put \( y = vx \)
Differentiating w.r.t. \( x \), \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
\( \therefore \) equation (i) becomes:
\( v + \frac{x d v}{d x} = \frac{x^2+v^2 x^2}{x^2+vx^2} \)
\( \implies v + \frac{x d v}{d x} = \frac{1+v^2}{1+v} \)
\( \implies \frac{x d v}{d x} = \frac{1+v^2}{1+v} - v \)
\( \implies \frac{x d v}{d x} = \frac{1+v^2-v-v^2}{1+v} \)
\( \implies \frac{x d v}{d x} = \frac{1-v}{1+v} \)
\( \implies \frac{x d v}{d x} = \frac{-(v-1)}{v+1} \)
\( \implies \frac{v+1}{v-1} dv = -\frac{dx}{x} \) (separately variables)
Interpreting both sides:
\( \int \frac{v+1}{v-1} dv = -\int \frac{dx}{x} \)
Adjustment:
\( \int \frac{v+1-1+1}{v-1} dv = -\log|x| \)
\( \implies \int \frac{(v-1)+2}{v-1} dv = -\log|x| \)
\( \implies \int \left( 1 + \frac{2}{v-1} \right) dv = -\log|x| \)
\( \implies v + 2\log|v - 1| = -\log|x| + c \)
Replace \( v \) by \( \frac{y}{x} \):
\( \implies \frac{y}{x} + 2\log\left|\frac{y}{x} - 1\right| + \log|x| = c \)
\( \implies \frac{y}{x} + \log\left|\left(\frac{y-x}{x}\right)^2 \cdot x\right| = c \)
Put \( x = 1 \) & \( y = 0 \):
\( \implies 0 + \log|1| = c \implies c = 0 \)
\( \therefore \frac{y}{x} + \log\left|\frac{(y-x)^2}{x}\right| = 0 \)
\( \implies \log\left|\frac{(y-x)^2}{x}\right| = -\frac{y}{x} \)
\( \implies \frac{(y-x)^2}{|x|} = e^{-\frac{y}{x}} \)
\( \implies (x - y)^2 = |x| e^{-\frac{y}{x}} \) is the required solution.

Question. Find the general solution: \( (x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy \).
Answer: \( \frac{dy}{dx} = \frac{x^3-3xy^2}{y^3-3x^2y} \)
Clearly the degree of each term in numerator & denominator is same.
It is a homogeneous D.E.
Put \( y = vx \)
Differentiating w.r.t. \( x \):
\( \frac{dy}{dx} = v + \frac{x d v}{d x} \)
\( \therefore \) equation (i) becomes:
\( v + \frac{x d v}{d x} = \frac{x^3-3v^2 x^3}{v^3 x^3-3v x^3} \)
\( \implies v + \frac{x d v}{d x} = \frac{1-3v^2}{v^3-3v} \)
\( \implies \frac{x d v}{d x} = \frac{1-3v^2}{v^3-3v} - v \)
\( \implies \frac{x d v}{d x} = \frac{-v^4-1}{v^3-3v} \)
\( \implies \frac{v^3-3v}{v^4-1} dv = -\frac{dx}{x} \)
Interpreting both sides:
\( \int \frac{v^3-3v}{v^4-1} dv = -\int \frac{dx}{x} \)
Separating:
\( \int \frac{v^3}{v^4-1} dv - 3 \int \frac{v}{v^4-1} dv = -\log|x| \)
Put \( v^4 - 1 = t \) and \( v^2 = z \) in 2nd integral:
\( v^3 dv = \frac{dt}{4} \) and \( vdv = \frac{dz}{2} \)
\( \implies \frac{1}{4} \int \frac{dt}{t} - \frac{3}{2} \int \frac{dz}{z^2-1} = -\log|x| \)
\( \implies \frac{1}{4}\log|t| - \frac{3}{2} \times \frac{1}{2} \log\left|\frac{z-1}{z+1}\right| = -\log|x| + \log c \)
\( \implies \frac{1}{4}\log\left|v^4 - 1\right| - \frac{3}{4}\log\left|\frac{v^2-1}{v^2+1}\right| = -\log|x| + \log c \)
\( \implies \log\left|v^4 - 1\right| - 3\log\left|\frac{v^2-1}{v^2+1}\right| = -4\log|x| + 4\log c \)
Replace \( v \) by \( \frac{y}{x} \):
\( \implies \log\left|\frac{y^4-x^4}{x^4}\right| - 3\log\left|\frac{y^2-x^2}{y^2+x^2}\right| + 4\log|x| = \log c^4 \)
\( \implies \log\left|\frac{\left(\frac{y^4-x^4}{x^4}\right) \cdot x^4}{\left(\frac{y^2-x^2}{y^2+x^2}\right)^3}\right| = \log c^4 \)
\( \implies \log\left|(y^4 - x^4) \cdot \left(\frac{y^2+x^2}{y^2-x^2}\right)^3\right| = \log c^4 \)
\( \implies (y^2 + x^2)(y^2 - x^2) \frac{(y^2+x^2)^3}{(y^2-x^2)^3} = c^4 \)
\( \implies \frac{(x^2+y^2)^4}{(x^2-y^2)^2} = c^4 \)
\( \implies (x^2 + y^2)^2 = (x^2 - y^2)c^2 \) is the required solution.

Question. Find the particular solution of the D.E. \( (3xy + y^2)dx + (x^2 + xy)dy = 0, x = 1 \& y = 1 \).
Answer: We have, \( (3xy + y^2)dx + (x^2 + xy)dy = 0 \)
\( \implies \frac{dy}{dx} = \frac{-(3xy+y^2)}{x^2+xy} \quad \dots (i) \)
It is a homogeneous D.E.
Put \( y = vx \)
Differentiating w.r.t. \( x \), \( \frac{dy}{dx} = v + \frac{x d v}{d x} \)
\( \therefore \) equation (i) becomes:
\( v + \frac{x d v}{d x} = \frac{-(3vx^2+v^2 x^2)}{x^2+vx^2} \)
\( \implies v + \frac{x d v}{d y} = \frac{-(3v+v^2)}{1+v} \)
\( \implies \frac{x d v}{d x} = \frac{-3v-v^2}{1+v} - v \)
\( \implies \frac{x d v}{d x} = \frac{-3v-v^2-v-v^2}{1+v} \)
\( \implies \frac{x d v}{d x} = \frac{-(2v^2+4v)}{1+v} \)
\( \implies \frac{v+1}{2v^2+4v} dv = -\frac{dx}{x} \) (separately variables)
Interpreting both sides:
\( \implies \int \frac{v+1}{2v^2+4v} dv = -\int \frac{dx}{x} \)
Put \( 2v^2 + 4v = t \)
\( (4v + 4)dv = dt \implies (v + 1)dv = \frac{dt}{4} \)
\( \therefore \frac{1}{4} \int \frac{dt}{t} = -\int \frac{dx}{x} \)
\( \frac{1}{4} \log|2v^2 + 4v| = -\log x + \log c \)
\( \implies \log|2v^2 + 4v| = -4\log x + 4\log c \)
Replace \( v \) by \( \frac{y}{x} \):
\( \implies \log\left|\frac{2y^2}{x^2} + \frac{4y}{x}\right| = \log \frac{c^4}{x^4} \)
\( \implies \log\left|\frac{2y^2+4xy}{x^2}\right| = \log\left(\frac{c^4}{x^4}\right) \)
\( \implies \frac{\left|2y^2+4xy\right|}{x^2} = \frac{c^4}{x^4} \)
\( \implies \left|4xy + 2y^2\right| = \frac{c_1}{x^2} \)

Question. Find one parameter solution of the D.E. \( x \cos\left(\frac{y}{x}\right) \cdot (ydx + xdy) = y \sin\left(\frac{y}{x}\right) \cdot (xdy - ydx) \).
Answer: \( xy \cos\left(\frac{y}{x}\right) dx + x^2 \cos\left(\frac{y}{x}\right) dy = xy \sin\left(\frac{y}{x}\right) dy - y^2 \sin\left(\frac{y}{x}\right) dx \)
\( \implies dy \left(x^2 \cos\left(\frac{y}{x}\right) - xy \sin\left(\frac{y}{x}\right)\right) = -dx \left(y^2 \sin\left(\frac{y}{x}\right) + xy \cos\left(\frac{y}{x}\right)\right) \)
\( \frac{dy}{dx} = \frac{-\left(y^2 \sin\left(\frac{y}{x}\right) + xy \cos\left(\frac{y}{x}\right)\right)}{x^2 \cos\left(\frac{y}{x}\right) - xy \sin\left(\frac{y}{x}\right)} \quad \dots (i) \)
It is a homogeneous D.E.
Put \( y = vx \)
Differentiating w.r.t. \( x \), \( \frac{dy}{dx} = v + \frac{x d v}{d y} \) put in eq. (i)
\( \implies v + \frac{x d v}{d x} = \frac{-v^2 x^2 \sin v + v x^2 \cos v}{x^2 \cos v - v x^2 \sin v} \)
\( \implies v + \frac{x d v}{d x} = \frac{-(v^2 \sin v + v \cos v)}{\cos v - v \sin v} \)
\( \implies \frac{x d v}{d x} = \frac{-v^2 \sin v - v \cos v}{\cos v - v \sin v} - v \)
\( \implies \frac{x d v}{d x} = \frac{-v^2 \sin v - v \cos v - v \cos v + v^2 \sin v}{\cos v - v \sin v} \)
\( \implies \frac{x d v}{d x} = \frac{-2v \cos v}{\cos v - v \sin v} \)
\( \implies \frac{\cos v - v \sin v}{v \cos v} = \frac{-2dx}{x} \)
Interpreting both sides:
\( \implies \int \frac{\cos v - v \sin v}{v \cos v} dv = -2 \int \frac{dx}{x} \)
\( \implies \int \left( \frac{1}{v} - \tan v \right) dv = -\log x \)
\( \implies \log v - \log \sec v = -\log x^2 + \log c \)
\( \implies \log \frac{v}{\sec v} = -\log x^2 + \log c \)
\( \implies \frac{xy}{\sec\left(\frac{y}{x}\right)} = c \)
\( \implies \log \left(\frac{y}{x}\right) \frac{x^2}{\sec\left(\frac{y}{x}\right)} = \log c \)
\( \implies \log \frac{xy}{\sec\left(\frac{y}{x}\right)} = \log c \)
\( \implies \frac{xy}{\sec\left(\frac{y}{x}\right)} = \pm c \)
\( \implies xy = c_1 \sec\left(\frac{y}{x}\right) \); where \( c_1 = \pm c \) is the required solution.

Question. Find the particular solution of the D.E. \( x \cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x \).
Answer: It is a homogeneous D.E.
Put \( y = vx \)
Differentiating w.r.t. \( x \), \( \frac{dy}{dx} = v + \frac{x d v}{d x} \)
\( \therefore v + \frac{x d v}{d x} = \frac{x+v x \cos v}{x \cos v} \)
\( \implies v + \frac{x d v}{d x} = \frac{1+v \cos v}{\cos v} \)
\( \implies \frac{x d v}{d x} = \frac{1+v \cos v}{\cos v} - v \)
\( \implies \frac{x d v}{d x} = \frac{1+v \cos v-v \cos v}{\cos v} \)
\( \implies \frac{x d v}{d x} = \frac{1}{\cos v} \)
\( \implies \cos v dv = \frac{dx}{x} \)
Interpreting both sides:
\( \int \cos v dv = \frac{dx}{x} \)
\( \implies \sin v = \log|x| + c \)
Replace \( v \) by \( \frac{y}{x} \):
\( \therefore \sin\left(\frac{y}{x}\right) = \log|x| + c \)
Put \( x = 1 \) and \( y = \frac{\pi}{4} \):
\( \implies \sin\left(\frac{\pi}{4}\right) = \log|1| + c \)
\( \implies \frac{1}{\sqrt{2}} = c \)
\( \therefore \sin\left(\frac{y}{x}\right) = \log|x| + \frac{1}{\sqrt{2}} \)

Question. Show that D.E. is homogeneous & solve it \( 2ye^{\frac{x}{y}}dx + \left(y - 2xe^{\frac{x}{y}}\right)dy = 0 \).
Answer: We have, \( 2ye^{\frac{x}{y}}dx = -\left(y - 2xe^{\frac{x}{y}}\right)dy = 0 \)
\( \implies \frac{dx}{dy} = \frac{2xe^{\frac{x}{y}}-y}{2ye^{\frac{x}{y}}} \quad \dots (i) \)
Here \( f(x, y) = \frac{2xe^{\frac{x}{y}}-y}{2ye^{\frac{x}{y}}} \)
\( f(\lambda x, \lambda y) = \frac{2\lambda xe^{\frac{\lambda x}{\lambda y}}-\lambda y}{2\lambda ye^{\frac{\lambda x}{\lambda y}}} \)
\( = \frac{\lambda}{\lambda} \left( \frac{2xe^{\frac{x}{y}}-y}{2ye^{\frac{x}{y}}} \right) \)
\( f(\lambda x, \lambda y) = \lambda^0 f(x, y) \)
Clearly function is homogeneous D.E.
Put \( x = vy \)
Differentiating w.r.t. \( y \), \( \frac{dx}{dy} = v + y\frac{dv}{dy} \)
\( \therefore \) equation (i) becomes:
\( v + y\frac{dv}{dy} = \frac{2vye^v-y}{2ye^v} \)
\( \implies v + y\frac{dv}{dy} = \frac{2vye^v-1}{2e^v} \)
\( \implies y\frac{dv}{dy} = \frac{2ve^v-1}{2e^v} - v \)
\( \implies y\frac{dv}{dy} = \frac{2ve^v-1-2ve^v}{2e^v} \)
\( \implies y\frac{dv}{dy} = -\frac{1}{2e^v} \)
\( \implies e^v dv = -\frac{1}{2} \frac{dy}{y} \)
Interpreting both sides:
\( \int e^v dv = -\frac{1}{2} \int \frac{dy}{y} \)
\( \implies e^v = -\frac{1}{2}\log|y| + c \)
Replace \( v \) by \( \frac{x}{y} \):
\( \implies e^{\frac{x}{y}} = -\frac{1}{2}\log|y| + c \) is the required solution.

Question. Solve the D.E. \( xe^{\frac{y}{x}} - y\sin\left(\frac{y}{x}\right) + x\frac{dy}{dx}\sin\left(\frac{y}{x}\right) = 0; y(1) = 0 \).
Answer: We have, \( xe^{\frac{y}{x}} - y\sin\left(\frac{y}{x}\right) + x\frac{dy}{dx}\sin\left(\frac{y}{x}\right) = 0 \)
\( \implies \frac{dy}{dx} = \frac{y\sin\left(\frac{y}{x}\right) - xe^{\frac{y}{x}}}{x\sin\left(\frac{y}{x}\right)} \quad \dots (i) \)
It is homogeneous D.E.
Put \( y = vx \)
Differentiating w.r.t. \( x \), \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) put in eq. (i)
\( \therefore v + \frac{x d v}{d x} = \frac{vx\sin v-xe^v}{x\sin v} \)
\( \implies v + \frac{x d v}{d x} = \frac{v\sin v-e^v}{\sin v} \)
\( \implies \frac{x d v}{d x} = \frac{v\sin v-e^v}{\sin v} - v \)
\( \implies \frac{x d v}{d x} = \frac{v\sin v-e^v-v\sin v}{\sin v} \)
\( \implies \frac{x d v}{d x} = \frac{-e^v}{\sin v} \)
\( \implies \frac{\sin v}{e^v} dv = \frac{-dx}{x} \)
\( \implies \int e^{-v} \sin v dv = -\int \frac{dx}{x} \quad \dots (ii) \)
Let \( I = \int e^{-v} \sin v dv \)
\( I = \sin v \cdot \frac{e^{-v}}{-1} - \int \cos v \cdot \frac{e^{-v}}{-1} dv \)
\( I = -e^{-v} \sin v + \int e^{-v} \cos v dv \)
\( I = -e^{-v} \sin v + \cos v \cdot \frac{e^{-v}}{-1} - \int -\sin v \cdot \frac{e^{-v}}{-1} dv \)
\( I = -e^{-v} \sin v - \cos v \cdot e^{-v} - I \)
\( 2I = -e^{-v}(\sin v + \cos v) \)
\( I = -\frac{e^{-v}}{2}(\sin v + \cos v) \)
\( \therefore \) equation (ii) becomes:
\( -\frac{e^{-v}}{2}(\sin v + \cos v) = -\log|x| + c \)
\( \implies e^{-v}(\sin v + \cos v) = 2\log|x| - 2c \)
Replace \( v \) by \( \frac{y}{x} \):
\( \implies e^{-\frac{y}{x}}\left(\sin\frac{y}{x} + \cos\frac{y}{x}\right) = \log\left|x^2\right| - 2c \)
Put \( x = 1 \) and \( y = 0 \):
\( \implies e^0(\sin 0 + \cos 0) = \log|1| - 2c \)
\( \implies 1 = -2c \implies c = -\frac{1}{2} \)
\( \therefore e^{-\frac{y}{x}}\left(\sin\frac{y}{x} + \cos\frac{y}{x}\right) = \log\left|x^2\right| + 1 \) is the required particular solution.

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