Question

Solve the differential equation  `ye^(x/y) dx = (xe^(x/y) + y^2)dy, (y != 0)`

Answer 1

`ye^(x/y)dx = (xe^(x/y) + y^2)dy`

`\implies ye^(x/y) dx/dy = xe^(x/y) + y^2`

`\implies e^(x/y) [y.dx/dy - x]` = y²

`\implies e^(x/y). ([y.dx/dy - x])/y^2` = 1              ...(1)

Let `e^(x/y)` = z.

Differentiating it with respect to y, we get:

`d/dy(e^(x/y)) = dz/dy`

`\implies e^(x/y) . d/dy (x/y) = dz/dy`

`\implies e^(x/y). [(y.dx/dy - x)/y^2] = dz/dy`              ...(2)

From equation (1) and equation (2), we get:

`dz/dy` = 1

`\implies` dz = dy

Integrating both sides, we get:

z = y + C

`\implies e^(x/y)` = y + C

Answer 2

`ye^(x/y)dx = (xe^(x/y) + y^2)dy`

`\implies e^(x/y) (ydx - xdy)` = y² dy

`\implies e^(x/y) ((ydx - xdy)/y^2)` = dy

`\implies e^(x/y)d(x/y)` = dy

`\implies int e^(x/y)d (x/y) = int dy`

`\implies e^(x/y)` = y + c, where 'c' is an arbitrary constant of integration.

Answer 3

We have, `dx/dy = (xe^(x/y) + y^2)/(y.e^(x/y)`

`\implies dx/dy = x/y + y/e^(x/y)` ...(i)

Let x = vy `\implies dx/dy = v + y.(dv)/dy`;

So equation (i) becomes `v + y (dv)/dy = v + y/e^v`

`\implies y (dv)/dy = y/e^v`

`\implies` e^v dv = dy

On integrating we get,

`inte^v dv = int dy`

`\implies` e^v = y + c

`\implies` e^x/y = y + c

where 'c' is an arbitrary constant of integration.