Question

Solve : dy/dx + y/x = y^2/x^2

Answer 1

Solution:

To solve differential equation: $\frac{dy}{dx} + \frac y x=\frac{y^2}{x^2}$

$\frac{dy}{dx} + \frac y x=\frac{y^2}{x^2} ...(i)$

Put $y=vx ...(ii)$

On differentiating w.r.t $x$

$\frac{dy}{dx}=v+\frac{xdv}{dx} ...(iii)$

Put (ii) and (iii) in (i), we get,

$v+\frac{xdv}{dx}+\frac{vx}{x}=\frac{v^2x^2}{x^2} \\ \Rightarrow v+\frac{xdv}{dx}+v=v^2 \\ \Rightarrow \frac{xdv}{dx}=v^2-2v \\ \Rightarrow \frac{dv}{v^2-2v}=\frac{dx}{x} \\ \Rightarrow \frac{dv}{(v-1)^2-1^2}=\frac{dx}{x}$

On integrating both sides,

$\frac 1 2 \ln |\frac{(v-1)-1}{(v-1)+1}|=\ln x+C \\ \Rightarrow \frac 1 2 \ln |\frac{v-2}{v}|=\ln x+C \\ \Rightarrow \ln |\frac{v-2}{v}|=2\ln x+2C$

$\\ \Rightarrow \ln |\frac{\frac y x-2}{\frac y x}|=2\ln x+2C$ [Using (ii)]

$\\ \Rightarrow \ln |\frac{y-2x}{y}|=\ln x^2+2C \\ \Rightarrow \ln |\frac{y-2x}{y}|-\ln x^2=2C \\ \Rightarrow \ln |\frac{y-2x}{yx^2}|=2C$

$\\ \Rightarrow\frac{y-2x}{yx^2}=e^{2C}$ [When $\ln y=x$, then $y=e^x$]

$\\ \Rightarrow\frac{y-2x}{yx^2}=k$ [where $k=e^{2C}$, constant]

$\Rightarrow y-2x=kyx^2 \\ \Rightarrow y-kyx^2=2x \\ \Rightarrow y(1-kx^2)=2x \\ \Rightarrow y=\frac{2x}{1-kx^2}$

Answer:

The solution of given differential equation is $y=\frac{2x}{1-kx^2}$ , where $k$ is a constant.