Question

Solve the differential equation $(x+y)dy+(x-y)dx=0$.

Answer 1

Consider $(x+y)dy+(x-y)dx=0$

$\Rightarrow \frac{dy}{dx}=-\frac{x-y}{x+y}...\left(1\right)$

It is a homogeneous function of degree $0$.

Put $y=vx$

Differentiating with respect to $x$ we get

$\frac{dy}{dx}=x\frac{dv}{dx}+v$

Thus $\left(1\right)$ becomes,

$x\frac{dv}{dx}+v=-\frac{x-vx}{x+vx}$

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

$\Rightarrow x\frac{dv}{dx}+v=-\frac{1-v}{1+v}$

$\Rightarrow x\frac{dv}{dx}=\frac{v-1}{1+v}-v$

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

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

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

Integrating both sides we get

$\Rightarrow \int \frac{1+v}{1+v^2}dv=-\int \frac{dx}{x}$

$\Rightarrow \int \frac{1}{1+v^2}dv+\int \frac{v}{1+v^2}dv=-\int \frac{dx}{x}$

$\Rightarrow ta{n}^{-1}\left(v\right)+\int \frac{v}{1+v^2}dv=-logx+logc$

Put $v^2+1=t\Rightarrow 2vdv=dt\Rightarrow vdv=\frac{dt}{2}$

$\therefore ta{n}^{-1}\left(v\right)+\int \frac{1}{2t}dt=-logx+logc$

$\Rightarrow ta{n}^{-1}\left(v\right)+\frac{1}{2}logt=-logx+logc$

Substitute back $t=v^2+1$

$\Rightarrow ta{n}^{-1}\left(v\right)+\frac{1}{2}log(v^2+1)=-logx+logc$

Substitute back $v=\frac{y}{x}$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+\frac{1}{2}log(\frac{y^2}{x^2}+1)=-logx+logc$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+\frac{1}{2}log\left(\frac{y^2+x^2}{x^2}\right)=-logx+logc$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+log\sqrt{\left(\frac{y^2+x^2}{x^2}\right)}=-logx+logc$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+log\left(\frac{\sqrt{y^2+x^2}}{x}\right)=-logx+logc$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+log\left(\sqrt{x^2+y^2}\right)-logx=-logx+logc$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+log\left(\sqrt{x^2+y^2}\right)=logc$