Question

Find the particular solution of the differential equation `(xe^(y/x) +y) dx = x dy,` given that y = 1 when x = 1.

Answer 1

`(xe^(y/x) +y) dx = x dy`

given y = 1, x = 1

The given differential eqn is homogeneous function of degree zero.

To solve it we make substitution

y = vx, `dy/dx = v+x (dv)/dx`   ...(i)

⇒ `dy/dx = (xe^(y/x)+y)/x`

⇒ `dy/dx = e^(y/x)+y/x`

from (i)

`v + x (dv)/dx = e^v + v`

⇒ `int (dv)/e^v = int dx/x`

−e^-v = log x + C or

`-e^(y/x) = log |x| + C`

It is given that x = 1, y = 1

So −e⁻¹ = log(1) + C

or C = −e⁻¹

Hence required solution

`e^-1 -e^(-y/x) = log |x|`