Question

Show that the differential equation 2y^x/y dx + (y − 2x e^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

${\displaystyle \frac{dx}{dy}=\frac{2x{e}^{\frac{x}{y}}-y}{2y{e}^{\frac{x}{y}}}.....................\left(1\right)}$

${\displaystyle LetF(x,y)=\frac{2x{e}^{\frac{x}{y}}-y}{2y{e}^{\frac{x}{y}}}}$

then ${\displaystyle F(\lambda x,\lambda y)=\frac{\lambda (2x{e}^{\frac{x}{y}}-y)}{\lambda \left(2y{e}^{\frac{x}{y}}\right)}={\lambda }^{\circ }\left[F(x,y)\right]}$

Thus, F (x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.

For solving, let us substitute x=vy     ..................(2)

Differentiating equation (2) with respect to y, we get

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

Substituting the value of x and  ${\displaystyle \frac{dx}{dy}}$ in equation (1), we get

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

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

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

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

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

${\displaystyle \text{or}2e^v=-\log \left|y\right|+C}$

Replacing v by x/y , we get

${\displaystyle 2e^{\frac{x}{y}}+\log \left|y\right|=c......\left(3\right)}$

Substituting x = 0 and y = 1in equation (3), we get

${\displaystyle 2e^0+\log \left|1\right|=c\Rightarrow c=2}$

Substituting the value of C in equation (3), we get

${\displaystyle 2e^{\frac{x}{y}}+\log \left|y\right|=2}$ .which is the particular solution of the given differential equation.