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.