Question

Find the particular solution of the differential equation ${\displaystyle (x{e}^{\frac{y}{x}}+y)dx=xdy,}$ given that y = 1 when x = 1.

Answer 1

${\displaystyle (x{e}^{\frac{y}{x}}+y)dx=xdy}$

given y = 1, x = 1

The given differential eqn is homogeneous function of degree zero.

To solve it we make substitution

y = vx, ${\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}}$   ...(i)

⇒ ${\displaystyle \frac{dy}{dx}=\frac{x{e}^{\frac{y}{x}}+y}{x}}$

⇒ ${\displaystyle \frac{dy}{dx}=e^{\frac{y}{x}}+\frac{y}{x}}$

from (i)

${\displaystyle v+x\frac{dv}{dx}=e^v+v}$

⇒ ${\displaystyle \int \frac{dv}{e^v}=\int \frac{dx}{x}}$

−e^-v = log x + C or

${\displaystyle -e^{\frac{y}{x}}=\log \left|x\right|+C}$

It is given that x = 1, y = 1

So −e⁻¹ = log(1) + C

or C = −e⁻¹

Hence required solution

${\displaystyle e^{-1}-e^{-\frac{y}{x}}=\log \left|x\right|}$