Question

Show that the given differential equation is homogeneous and solve:\(y\,dx+xlog(\frac{y}{x})dy-2x\,dy=0\)

Answer 1

\(y\,dx+xlog(\frac{y}{x})dy-2x\,dy=0\)
\(⇒y\,dx=[2x-xlog(\frac{y}{x})]dy\)
\(⇒\frac{dy}{dx}=\frac{y}{2x-xlog(\frac{y}{x})}....(1)\)
Let \(F(x,y)=\frac{y}{2x-xlog(\frac{y}{x})}.\)
\(∴F(λx,λy)=\frac{λy}{2(λx)-(λx)log(\frac{λy}{λx})}=\frac{y}{2x-log(\frac{y}{x})}=λ.F(x,y)\)
Therefore,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
\(y=vx\)
\(⇒\frac{dy}{dx}=\frac{d}{dx}(vx)\)
\(⇒\frac{dy}{dx}=v+x\frac{dv}{dx}\)
Substituting the values of y and \(\frac{dy}{dx}\) in equation(1),we get:
\(v+x\frac{dv}{dx}=\frac{vx}{2x-xlogv}\)
\(⇒v+x\frac{dv}{dx}=\frac{v}{2-logv}\)
\(⇒x\frac{dv}{dx}=\frac{v}{2-logv}-v\)
\(⇒x\frac{dv}{dx}=\frac{v-2v+v\,logv}{2-logv}\)
\(⇒x\frac{dv}{dx}=\frac{v\,logv-v}{2-logv}\)
\(⇒\frac{2-logv}{v(logv-1)}dv=\frac{dx}{x}\)
\(⇒[\frac{(1+(1-logv)}{v(logv-1)}]dv=\frac{dx}{x}\)
\(⇒[\frac{1}{v(logv-1)}-\frac{1}{v}]dv=\frac{dx}{x}\)
Integrating both sides,we get:
\(∫\frac{1}{v(logv-1)}dv-∫\frac{1}{v}dv=∫\frac{1}{x}dx\)
\(⇒∫\frac{dv}{v(logv-1)}-logv=logx+logC...(2)\)
\(⇒Let\,\, log\,v-1=t\)
\(⇒\frac{d}{dv}=(logv-1)=\frac{dt}{dv}\)
\(⇒\frac{1}{v}=\frac{dt}{dv}\)
\(⇒\frac{dv}{v}=dt\)
Therefore,equation(1),becomes:
\(⇒∫\frac{dt}{t}-logv=logx+logC\)
\(⇒logt-log(\frac{y}{x})=log(Cx)\)
\(⇒log[log(\frac{y}{x})-1]-log(\frac{y}{x})=(Cx)\)
\(⇒log[\frac{log(\frac{y}{x})-1}{\frac{y}{x}}]=log(Cx)\)
\(⇒\frac{x}{y}[log(\frac{y}{x})-1]=Cx\)
\(⇒log(\frac{y}{x})-1=Cy\)
This is the required solution of the given differential equation.