Question

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

• A. $y{e}^{\frac{y}{x}}+x=c$
• B. $y{e}^{\frac{x}{y}}-x=c$
• C. $y{e}^{\frac{x}{y}}+y=c$
• D. $y{e}^{\frac{x}{y}}+x=c$

Answer 1

Answer: (D)

$y{e}^{\frac{x}{y}}+x=c$

The given differential equation is,

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

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

$\Rightarrow$ Let $x=vy$

$\Rightarrow$ $\frac{dx}{dy}=v+y\frac{dv}{dy}$

$v+y\frac{dv}{dy}=\frac{e^v(v-1)}{1+e^v}$

$\Rightarrow$ $y\frac{dv}{dy}=\frac{v{e}^v-e^v}{1+e^v}-v$

$\Rightarrow$ $y\frac{dv}{dy}=\frac{v{e}^v-e^v-v-v{e}^v}{1+e^v}$

$\Rightarrow$ $y\frac{dv}{dy}=\frac{-(v+e^v)}{(1+e^v)}$

$\Rightarrow$ $\frac{(1+e^v)}{(v+e^v)}dv=-\frac{dy}{y}$

Integrating both sides, we get

$\int \frac{(1+e^v)}{(v+e^v)}dv=-\int \frac{dy}{y}$ ------ ( 2 )

Let $I=\int \frac{(1+e^v)}{(v+e^v)}dv$

Put $v+e^v=t$

$(1+e^v)dv=dt$

So, $I=\int \frac{dt}{t}=\log t=\log (v+e^v)$

So, from ( 2 ), we get

$\log (v+e^v)=-\log y+\log C$

$\Rightarrow$ $\log (\frac{x}{y}+e^{\frac{x}{y}})+\log y=\log C$

$\Rightarrow$ $\log \left[y(\frac{x}{y}+e^{\frac{x}{y}})\right]=\log C$

$\Rightarrow$ $y(\frac{x}{y}+e^{\frac{x}{y}})=C$

$\Rightarrow$ $x+y{e}^{\frac{x}{y}}=C$