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.