Comment
Share
Q)

# Solve the differential equation$y\;e^{\large\frac{x}{y}}\;dx=\bigg(x\;e^{\Large\frac{x}{y}}+y^2\bigg)dy\;(y\neq0)$

Comment
A)
Toolbox:
• If the linear differential equation is of the form $\large\frac{dx}{dy}$$= F(x,y), is said to be homogenous if F(x,y) is a homogenous function of degree 0. • This type of equation can be solved by substituting x= vy and \large\frac{dx}{dy} =$$ v+y\large\frac{dv}{dy}$
Step 1:
Given :$y\;e^{\large\frac{x}{y}}\;dx=\bigg(x\;e^\frac{x}{y}+y^2\bigg)dy$
we can write the equation as $\large\frac{dx}{dy} = \frac{[xe^{\Large(x/y)} +y^2]}{y.e^{\Large(x/y)} }$
Using the information in the tool box, we understand this given equation is a homogenous function of dergree 0,
Step 2:
let us substitute $x= vy$ and $\large\frac{dx}{dy}$$= v + y\large\frac{dv}{dy}$
$v + y\large\frac{dv}{dy} =\large\frac{ [vy.e^v + y^2]}{y.e^v}$
taking y as the common factor and cancelling we get,
$v + y\large\frac{dv}{dx} =\frac{ [ve^y + y]}{e^v}$
bringing v from LHS to RHS we get
$y\large\frac{dv}{dy} =\frac{ y}{e^v }$
seperating the variables we get,
$e^vdv = dy$
Step 3:
Integrating on both sides we get,
$\int e^vdv=\int dy$
$e^v=y+C$
Substituting for $v=\large\frac{ x}{y}$ we get,
$e^{\Large\frac{x}{y}}=y+C$
This is the required solution.