Question

Show that the given differential equation is homogeneous and solve:\(x\frac{dy}{dx}-y+sinx(\frac{y}{x})=0\)

Answer 1

\(x\frac{dy}{dx}-y+sinx(\frac{y}{x})=0\)
\(⇒x\frac{dy}{dx}=y-xsin(\frac{y}{x})\)
\(⇒\frac{dy}{dx}=\frac{y-xsin(\frac{y}{x})}{x}...(1)\)
Let \(F(x,y)=\frac{y-xsin(\frac{y}{x})}{x}.\)
\(∴F(λx,λy)=\frac{λy-λxsin(\frac{λy}{λx})}{λx}=\frac{y-sinx(\frac{y}{x})}{x}=λ.F(x,y)\)
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 y and \(\frac{dy}{dx}\) in equation(1),we get:
\(v+x\frac{dv}{dx}=\frac{vx-xsinv}{x}\)
\(⇒v+x\frac{dv}{dx}=v-sinv\)
\(⇒-\frac{dv}{sinv}=\frac{dx}{x}\)
\(⇒cosec\,v\,\, dv=\frac{-dx}{x}\)
Integrating both sides,we get:
\(log|cosec\,v-cotv|=-logx+logC=log \frac{C}{x}\)
\(⇒cosec(\frac{y}{x})-cot(\frac{y}{x})=\frac{C}{x}\)
\(⇒\frac{1}{sin(\frac{y}{x})}-\frac{cos(\frac{y}{x})}{sin(\frac{y}{x})}=\frac{C}{x}\)
\(⇒x[1-cos(\frac{y}{x})]=Csin(\frac{y}{x})\)
This is the required solution of the given differential equation.