Question

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

Answer 1

The given differential equation is:
${xcos(\frac{y}{x})+ysin(\frac{y}{x})}y\,dx={ysin(\frac{y}{x})-xcos(\frac{y}{x})}x\,dy$
$\frac{dy}{dx}=\frac{{xcos(\frac{y}{x})+ysin(\frac{y}{x})}y}{{ysin(\frac{y}{x})-xcos(\frac{y}{x})}x}...(1)$
Let $F(x,y)=\frac{{xcos(\frac{y}{x})+ysin(\frac{y}{x})}y}{{ysin(\frac{y}{x})-xcos(\frac{y}{x})}x}$
∴F(λx,λy)=\frac{{λxcos(\frac{λy}{λx})+λysin(\frac{λy}{λx})}λy}{{λysin(\frac{λy}{λx})-λxsin(\frac{λy}{λx})}λx}$$=\frac{{xcos(\frac{y}{x})+ysin(\frac{y}{x})}y}{{ysin(\frac{y}{x})-xcos(\frac{y}{x})}x}$$=λ.F(x,y)
Therefore,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
$y=vx$
$⇒\frac{dy}{dx}=v+x=\frac{dy}{dx}$
Substituting the values of y and $\frac{dy}{dx}$ in equation(1),we get:
$v+x\frac{dv}{dx}=\frac{(xcosv+vx sinv).vx}{(vx sinv-xcosv).x}$
$⇒v+x\frac{dv}{dx}=\frac{vcosv+v^2sinv}{vsinv-cosv}$
$⇒x\frac{dv}{dx}=\frac{vcosv+v^2sinv}{vsinv-cosv}-v$
$⇒x\frac{dv}{dx}=\frac{vcosv+v^2sinv-v^2sinv+vcosv}{vsinv-cosv}$
$⇒x\frac{dv}{dx}=\frac{2cosv}{vsinv-cosv}$
$⇒[\frac{vsinv-cosv}{vcosv}]dv=2\frac{dx}{x}$
$⇒(\frac{tan^{-1}v}{v})dv=2\frac{dx}{x}$
Integrating both sides,we get:
$log(secv)-logv=2logx+logC$
$⇒log(\frac{secv}{v})=log(Cx)^2$
$⇒(\frac{secv}{v})=Cx^2$
$⇒secv=Cx^2v$
$⇒sec(\frac{y}{x})=C.x^2.\frac{y}{x}$
$⇒sec(\frac{y}{x})=Cxy$
$⇒cos(\frac{y}{x})=\frac{1}{cxy}=\frac{1}{C}.\frac{1}{xy}$
$⇒xy\, cos(\frac{y}{x})=k\,\,\ (k=\frac{1}{C})$
This is the required solution of the given differential equation.