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Mathematics Question on Differential equations

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

Answer

(1+exy)dx+exy(1−xy)dy=0(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0
(1+exy)dx=−exy(1−xy)dy=0(1+e^{\frac{x}{y}})dx=-e^{\frac{x}{y}}(1-\frac{x}{y})dy=0
⇒dxdy=−exy(1−xy)1+exy...(1)⇒\frac{dx}{dy}=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}...(1)
Let F(x,y)=−exy(1−xy)1+exyF(x,y)=\frac{-e^{\frac{x}{y}}(1-\frac{x}{y})}{1+e^{\frac{x}{y}}}
∴F(λx,λy)=−eλxλy(1−λxλy)1+eλxλy=−exy(1−xy)1+exy=λ0.F(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=vyx=vy
⇒ddy(x)=ddy(vy)⇒\frac{d}{dy}(x)=\frac{d}{dy}(vy)
⇒dxdy=v+ydvdy⇒\frac{dx}{dy}=v+y\frac{dv}{dy}
Substituting the values of x and dxdy\frac{dx}{dy} in equation(1),we get:
v+ydvdy=−ev(1−v)1+evv+y\frac{dv}{dy}=\frac{-e^v(1-v)}{1+e^v}
⇒ydvdy=−ev+vev1+ev−v⇒y\frac{dv}{dy}=\frac{-e^v+ve^v}{1+e^v}-v
⇒ydvdy=−ev+vev−v−vev1+ev⇒y\frac{dv}{dy}=\frac{-e^v+ve^v-v-ve^v}{1+e^v}
⇒ydvdy=−[v+ev1+ev]⇒y\frac{dv}{dy}=-[\frac{v+e^v}{1+e^v}]
⇒[1+evv+ev]dv=−dyy⇒[\frac{1+e^v}{v+e^v}]dv=\frac{-dy}{y}
Integrating both sides,we get:
⇒log(v+ev)=−logy+logC=log(Cy)⇒log(v+e^v)=-logy+logC=log(\frac{C}{y})
⇒[xy+exy]=Cy⇒[\frac{x}{y}+e^{\frac{x}{y}}]=\frac{C}{y}
⇒x+yexy=C⇒x+ye^{\frac{x}{y}}=C
This is the required solution of the given differential equation.