Question

Show that the differential equation
$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$ is homogeneous and find its particular solution, given that $y=1$ at $x=0$

Answer 1

Lets take the given differential equation
$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0$

It can be written as
$\frac{dx}{dy}=\frac{e^{\frac{x}{y}}(\frac{x}{y}-1)}{(1+e^{\frac{x}{y}})}...\left(1\right)$

R.H.S. is a function of $x$ and $y$
$\therefore \frac{dx}{dy}=f(x,y)=\frac{e^{\frac{x}{y}}(\frac{x}{y}-1)}{(1+e^{\frac{x}{y}})}$

Put $x=\lambda x$ and $y=\lambda y$, we get
$f(\lambda x,\lambda y)=\frac{e^{\frac{\lambda x}{\lambda y}}(\frac{\lambda x}{\lambda y}-1)}{(1+e^{\frac{\lambda x}{\lambda y}})}$
$\Rightarrow f(\lambda x,\lambda y)={\lambda }^0\left[\frac{e^{\frac{x}{y}}(\frac{x}{y}-1)}{(1+e^{\frac{x}{y}})}\right]$

$\Rightarrow f(\lambda x,\lambda y)={\lambda }^{0}f(x,y)$

$\therefore f(x,y)$ is a homogeneous function of degree zero.

Hence, the given differential equation is homogeneous differential equation.

To solve this homogeneous differential equation,
Put $x=vy$
Differentiate on both sides,
$\frac{dx}{dy}=v+y\frac{dv}{dy}$

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

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

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

On saparating the variables,
$\frac{(1+e^v)}{(v+e^v)}dv=-\frac{dy}{y}$

Integrate on both sides,
$\int \frac{(1+e^v)}{(v+e^v)}dv=-\int \frac{dy}{y}$
$\Rightarrow \log (v+e^v)=-\log y+\log C(\because \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log (f\left(x\right)))$
$\Rightarrow \log (e^v+v)=\log \frac{C}{y}$

Take anti-log on both sides,
$\Rightarrow (e^v+v)=\frac{C}{y}$
$\Rightarrow y(e^{\frac{x}{y}}+\frac{x}{y})=C$

To find the value of C,
put $x=0$ and $y=1$ we get, $C=1$.
$\therefore x+y{e}^{\frac{x}{y}}=1$