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.