Question

Show that the differential equation $(x{e}^{x/y}+y)dx=xdy$ is homogeneous. Find the particular solution of this differential equation, given that $x=1$ when $y=1$.

Answer 1

The given differential equation can be written as

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

Let $F(x,y)=\frac{x}{x{e}^{\frac{x}{y}}+y}$
Then $F(\lambda x,\lambda y)=\frac{\lambda x}{\lambda (x{e}^{\frac{x}{y}}+y)}={\lambda }^o\left[F\right(x,y\left)\right]$
Thus, $F(x,y)$ is a homogeneous function of degree zero.

Therefore, the given differential equation is a homogeneous differential equation.

To solve it, we make the substitution
$x=vy$ --- (2)

Differentiating equation (1) and (2) w.r.t to $y$, we get
$\frac{dx}{dy}=v+y\frac{dv}{dy}$

Substitute the value of $x$ and $\frac{dx}{dy}$ in equation (1), we get
$v+y\frac{dv}{dy}=\frac{v}{v{e}^v+1}$

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

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

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

Integrating on both the sides, we get
$\int \frac{v{e}^v+1}{v^2{e}^v}dv=-\int \frac{dy}{y}$

$\int \frac{v{e}^v}{v^2{e}^v}dv+\frac{1}{v^2{e}^v}dv=-\log \left|y\right|+C$

$\log \left|v\right|-\frac{1}{v{e}^v}[\frac{1}{v}+1]=-\log \left|y\right|+C$

Replacing v by $\frac{x}{y}$
$\log \left|\frac{x}{y}\right|-\frac{1}{\frac{x}{y}{e}^{\frac{x}{y}}}[\frac{1}{\frac{x}{y}}+1]=-\log \left|y\right|+C$-----(3)

Substituting x = 1 and y =1 in equation 3, we get
$\log \left|\frac{1}{1}\right|-\frac{1}{\frac{1}{1}{e}^{\frac{1}{1}}}[\frac{1}{\frac{1}{1}}+1]=-\log \left|1\right|+C\Rightarrow C\approx -0.7$

Therefore, the solution is

$\log \left|\frac{x}{y}\right|-\frac{1}{\frac{x}{y}{e}^{\frac{x}{y}}}[\frac{1}{\frac{x}{y}}+1]=-\log \left|y\right|-0.7$