Question

Solve the differential equation :
$e^{\frac{x}{y}}\left(\begin{array}{c}1-\frac{x}{y}\end{array}\right)+(1+e^{\frac{x}{y}})\frac{dx}{dy}=0$ when $x=0,y=1$

Answer 1

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

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

Since (i) is of the form $\frac{dx}{dy}=g\left(\frac{x}{y}\right)$. It is a homogeneous differential equation.

Put $x=vy$

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

Substituting the values of $x$ and $\frac{dx}{dy}$ in (i), we get

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

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

$=\frac{-v-e^v}{1+e^v}$

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

On integrating both sides, we get

$\log |v+e^v|=-\log |y|+c$

$\Rightarrow \log |y(v+e^v)|=c$

$\Rightarrow y\left(\begin{array}{c}\frac{x}{y}+e^{\frac{x}{y}}\end{array}\right)=\pm e^c$

$\Rightarrow x+y{e}^{\frac{x}{y}}=A$ ...(i)

Where, A is arbitrary constant.

When $x=0,y=1,$ $A=1$ $[\because 0+1e^{0/1}=0+1\times 1=1]$

Substituting this value of A in equation (i), we get

$x+y{e}^{\frac{x}{y}}=1$, which is the required particular solution.