Question

Show that the given differential equation is homogeneous and solve:\((1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0\)

Answer 1

\((1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0\)
\((1+e^{\frac{x}{y}})dx=-e^{\frac{x}{y}}(1-\frac{x}{y})dy=0\)
\(⇒\frac{dx}{dy}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}...(1)\)
Let \(F(x,y)=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}\)
\(∴F(λx,λy)=\frac{-e^{\frac{λx}{λy}}(1-\frac{λx}{λy})}{1+e^{\frac{λx}{λy}}}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}=λ0.F(x,y)\)
Therefore,the given differential equation is a homogenous equation.
To solve it,we make the substitution as:
\(x=vy\)
\(⇒\frac{d}{dy}(x)=\frac{d}{dy}(vy)\)
\(⇒\frac{dx}{dy}=v+y\frac{dv}{dy}\)
Substituting the values of x and \(\frac{dx}{dy}\) in equation(1),we get:
\(v+y\frac{dv}{dy}=\frac{-e^v(1-v)}{1+e^v}\)
\(⇒y\frac{dv}{dy}=\frac{-e^v+ve^v}{1+e^v}-v\)
\(⇒y\frac{dv}{dy}=\frac{-e^v+ve^v-v-ve^v}{1+e^v}\)
\(⇒y\frac{dv}{dy}=-[\frac{v+e^v}{1+e^v}]\)
\(⇒[\frac{1+e^v}{v+e^v}]dv=\frac{-dy}{y}\)
Integrating both sides,we get:
\(⇒log(v+e^v)=-logy+logC=log(\frac{C}{y})\)
\(⇒[\frac{x}{y}+e^{\frac{x}{y}}]=\frac{C}{y}\)
\(⇒x+ye^{\frac{x}{y}}=C\)
This is the required solution of the given differential equation.