Question

Particular solution of the differential equation \(\frac{dy}{dx} = \frac{y^2-2xy-x^2}{y^2+2xy-x^2}\)  given y=-1 at x=1.

(a) y=x

(b) y+x=2

(c) y=-x

(d) y-x=2

I have been asked this question in an online interview.

This is a very interesting question from Separable and Homogeneous Equations in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

The correct answer is (c) y=-x

Easy explanation: \(\frac{dy}{dx} = \frac{y^2-2xy-x^2}{y^2+2xy-x^2} = \frac{\frac{y^2}{x^2} – \frac{2y}{x} – 1}{\frac{y^2}{x^2} + \frac{2y}{x} -1} \)……. is a homogeneous equation

thus put \(y = vx \rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} = \frac{v^2-2v-1}{v^2+2v-1}\)

separating the variables and integrating we get

\(x\frac{dv}{dx} = \frac{v^2-2v-1}{v^2+2v-1} – v = -\frac{(v^3+v^2+v+1)}{v^2+2v-1} = -\frac{(v^2+1)(v+1)}{v^2+2v-1}\)

\(\int \frac{v^2+2v-1}{(v^2+1)(v+1)} \,dv = \int \frac{-1}{x} \,dx\)

\(\int\frac{2v(v+1)-(v^2+1)}{(v^2+1)(v+1)} \,dv = log \,c – log \,x\)

\(\int(\frac{2v}{v^2+1}-\frac{1}{v+1}) \,dv = log \,c – log \,x\)

log(v^2+1) – log(v+1) + log x = log c –> \(\frac{(v^2+1)x}{(v+1)}=c\)

\(\rightarrow \frac{x^2+y^2}{x+y} = c \rightarrow k(x^2+y^2)=(x+y)\) where k=1/c

at x=1, y=-1 substituting we get 2k=0→k=0

thus the particular solution is y=-x.