Question

Show that the given differential equation is homogeneous and then solve it.

$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$

Answer 1

Given, $(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$
$\Rightarrow \frac{dx}{dy}=\frac{e^{x/y}(\frac{x}{y}-1)}{e^{x/y}+1}$ ...(i)
Thus, the given differential equation is homogeneous
So, put $\frac{x}{y}=v$ i.e., $x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$
Then, Eq. (i) becomes
$v+y\frac{dv}{dy}=\frac{e^v(v-1)}{e^v+1}\Rightarrow y\frac{dv}{dy}=\frac{e^v(v-1)}{e^v+1}-v\Rightarrow y\frac{dv}{dy}=\frac{v{e}^v-e^v-v{e}^v-v}{e^v+1}\Rightarrow \frac{e^v+1}{e^v+v}dv=-\frac{1}{y}dy$
On integrating both sides, we get
$\int \frac{e^v+1}{e^v+v}dv=-\int \frac{1}{y}dyPut{e}^v+v=t\Rightarrow e^v+1=\frac{dt}{dv}\Rightarrow dv=\frac{dt}{e^v+1}\therefore \int \frac{e^v+1}{t}\frac{dt}{e^v+1}=-log|y|+logC\Rightarrow log|t|+log|y|=logC\Rightarrow log|e^v+v|+log|y|=logC$ $[\because t=e^v+v]$
$\Rightarrow log|(e^v+v)y|=logC\Rightarrow |(e^v+y)|=C\Rightarrow (e^v+v)y=C$
So, put $\frac{x}{y}=v$, we get
$(e^{x/y}+\frac{x}{y})y=C\Rightarrow y{e}^{x/y}+x=C$
This is the required solution of the given differential equation.