Question

Show that the given differential equation is homogeneous and then solve it.

(x-y)dy-(x+y)dx=0.

Answer 1

Given, $\Rightarrow (x-y)dy=(x+y)dx\Rightarrow \frac{dy}{dx}=\frac{x+y}{x-y}$ ...(i)
Thus, the given differential equation is homogeneous.
So put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
On putting values of $\frac{dy}{dx}$ and y in Eq.(i), we get
$v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}\Rightarrow v+x\frac{dv}{dx}=\frac{1+v}{1-v}\Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v\Rightarrow x\frac{dv}{dx}=\frac{1+v-v+v^2}{1-v}\Rightarrow \frac{1-v}{1+v^2}dv=\frac{dx}{x}$
On integrating both sides, we get
$\int \frac{1-v}{1+v^2}dv=\int \frac{dx}{x}\Rightarrow \int \frac{1}{1+v^2}dv-\int \frac{v}{1+v^2}dx=log|x|+CLet1+v^2=t\Rightarrow 2v=\frac{dt}{dv}\Rightarrow dv=\frac{dt}{2v^{\prime }}\therefore ta{n}^{-1}v-\int \frac{v}{t}\times \frac{dt}{2v}=log|x|+C\Rightarrow ta{n}^{-1}v-\frac{1}{2}log|t|=log|x|+C\Rightarrow 2ta{n}^{-1}v-\left[log\right(1+v^2)+2log(x\left)\right]=2C$ $(Putt=1+v^2)$
$\Rightarrow 2ta{n}^{-1}v-log\left[\right(1+v^2\left)x^2\right]=2C$
$\Rightarrow 2ta{n}^{-1}\frac{y}{x}-log\left[\left(\frac{x^2+y^2}{x^2}\right)x^2\right]=2C$ $(Putv=\frac{y}{x})$
$\Rightarrow 2ta{n}^{-1}\frac{y}{x}-log(x^2+y^2)=2C\Rightarrow ta{n}^{-1}\frac{y}{x}-\frac{1}{2}log(x^2+y^2)=C$
This is the required solution of the given differential equation.