Question

Show that the given differential equation is homogeneous and solve:$x\,dy-y\,dx=\sqrt{x^2+y^2}\,dx.$

Answer 1

$x\,dy-y\,dx=\sqrt{x^2+y^2}\,dx.$
$⇒xdy=[y+\sqrt{x^2+y^2}]dx$
$\frac{dy}{dx}=y+\sqrt{x^2+\frac{y^2}{x^2}}...(1)$
Let $F(x,y)=y+\sqrt{x^2+\frac{y^2}{x^2}}$
$∴F(λx,λy)=λx+\sqrt{(λx)^2+\frac{(λy)^2}{λx}}=y+\sqrt{x^2+\frac{y^2}{x}}=λ.F(x,y)$
Therefore,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
$y=vx$
$⇒\frac{d}{dx}(y)=\frac{d}{dx}(vx)$
$⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$
Substituting the values of v and $\frac{dy}{dx}$ in equation(1),we get:
$v+x\frac{dv}{dx}=vx+\sqrt{x^2+\frac{(vx)^2}{x}}$
$⇒v+x\frac{dv}{dx}=v+\sqrt{1+v^2}$
$⇒\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}$
Integrating both sides,we get:
$log|v+\sqrt{1+v^2}|=log|x|+logC$
$⇒log|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}|=log|Cx|$
$⇒log|\frac{y+\sqrt{x^2+y^2}}{x}|=log|Cx|$
$⇒y+\sqrt{x^2+y^2}=Cx^2$
This is the required solution of the given differential equation.