Question

Show that the differential equation $xdy-ydx=\sqrt{x^2+y^2}dx$ is homogeneous and solve it.

Answer 1

Given differential equation:

$xdy-ydx=\sqrt{x^2+y^2}dx$

$xdy=(\sqrt{x^2+y^2}+y)dx$

$\frac{dy}{dx}=\frac{\sqrt{x^2+y^2}+y}{x}$

If $F(\lambda x,\lambda y)=\lambda F(x,y)$, then the differential equation is homogeneous.

Substituting $F(x,y)=\frac{\sqrt{x^2+y^2}+y}{x}$

Now,

$F(\lambda x,\lambda y)=\frac{\sqrt{(\lambda x{)}^2+(\lambda y{)}^2}+\lambda y}{\lambda x}$

$F(\lambda x,\lambda y)=\frac{\sqrt{{\lambda }^2{x}^2+{\lambda }^2{y}^2}+\lambda y}{\lambda x}$

$F(\lambda x,\lambda y)=\frac{\lambda (\sqrt{x^2+y^2}+y)}{\lambda x}$

$F(\lambda x,\lambda y)=\frac{\sqrt{x^2+y^2}+y}{x}=F(x,y)$

Hence, the differential equation is homogeneous.

Given differential equation:

$xdy-ydx=\sqrt{x^2+y^2}dx$

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

Let $y=vx$

Differentiating both sides $w.r.t.{}^{\prime }{x}^{\prime }$, we get

$\frac{dy}{dx}=\frac{d\left(vx\right)}{dx}$

Using product rule : $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

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

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

Substituting the value of $\frac{dy}{dx}\text{and}y=vx$
in $\left(i\right)$, we get

$x\frac{dv}{dx}+v=\frac{\sqrt{x^2+(vx{)}^2}+\left(vx\right)}{x}$

$x\frac{dv}{dx}+v=\frac{\sqrt{x^2+x^2{v}^2}+vx}{x}$

$x\frac{dv}{dx}+v=\frac{x\sqrt{1+v^2}+vx}{x}$

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

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

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

Integrating both sides, we get

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

$\log |v+\sqrt{v^2+1}|=\log \left|x\right|+\log c$

$\log |v+\sqrt{v^2+1}|=\log \left|cx\right|$

$v+\sqrt{v^2+1}=cx$

Substituting $v=\frac{y}{x}$,

$\frac{y}{x}+\sqrt{{\left(\frac{y}{x}\right)}^2+1}=cx$

$\frac{y}{x}+\sqrt{\frac{y^2}{x^2}+1}=cx$

$\frac{y}{x}+\frac{\sqrt{y^2+x^2}}{x}=cx$

$y+\sqrt{y^2+x^2}=c{x}^2$

Hence, the required general solution is

$y+\sqrt{y^2+x^2}=c{x}^2$