Question

Show that the differential equation $2y{e}^{\frac{x}{y}}dx+(y-2x{e}^{\frac{x}{y}})dy=0$ is homogeneous. Find the particular solution of this differential equation, given that $x=0$, when $y=1$.

Answer 1

The given differential equation can be written as $\frac{dx}{dy}=\frac{2x{e}^{\frac{x}{y}-y}}{2y{e}^{\frac{x}{y}}}$ ---- (1)
Let $F(x,y)=\frac{2x{e}^{\frac{x}{y}-y}}{2y{e}^{\frac{x}{y}}}$
Then $F(\lambda x,\lambda y)=\frac{\lambda 2x{e}^{\frac{x}{y}-y}}{\lambda 2y{e}^{\frac{x}{y}}}={\lambda }^o\left[F\right(x,y\left)\right]$

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

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

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

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

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

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

$2e^v=-\log \left|y\right|+C$

Now replacing v by $\frac{x}{y}$, we get
$2e^{\frac{x}{y}}+\log \left|y\right|=C$ ----(3)

Substituting $x=0$ and $y=1$ in equation (3), we get
$2e^{\frac{0}{1}}+\log \left|1\right|=C\Rightarrow C=2$
$2e^{\frac{x}{y}}+\log \left|y\right|=2$