Browse Questions

# Solve the differential equation $y.e^{\large\frac{x}{y}} dx=(x.e^{\large\frac{x}{y}}+y^2)dy ; y \neq 0$

Toolbox:
• A differential equation of the form $\large\frac{dy}{dx }$$= F(x,y) is said to be homogenous if F(x,y) is a homogenous function of degree zero. • To solve this type of equations substitute y = vx and \large\frac{dy}{dx }$$= v + x\large\frac{dv}{dx}$
Step 1:
$ye^{\Large\frac{x}{y}}dx=(xe^{\Large\frac{x}{y}}+y^2)dy$
This can be written as
$\large\frac{dx}{dy}=\frac{xe^{\Large x/y}+y^2}{ye^{\Large x/y}}$
Step 2:
Clearly this is a homogeneous differential equation .
Hence put $x=vy$
$\large\frac{dx}{dy}=$$v+y\large\frac{dv}{dy} \therefore v+y\large\frac{dv}{dy}=\frac{vye^v+y^2}{ye^v} \Rightarrow \large\frac{ ve^v+y}{e^v} \Rightarrow y\large\frac{dv}{dy}=\frac{ve^v+y}{e^v}$$-v$
$y\large\frac{dv}{dy}=\frac{y}{e^v}$
Step 3:
Now separating the variables we get,
$e^vdv=dy$
On integrating we get
$\int e^vdv=\int dy$
(i.e) $e^v=y+c$
Substituting for $v=\large\frac{x}{y}$ we get,