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.