Question

The general solution of `(1 + e^{x/y})dx + e^{x/y}(1 - x/y)dy = 0` is ______ 

Options

• `ye^{y/x} + x = c`

• `ye^y - x = c`

• `ye^{x/y} + y = c`

• `ye^{x/y} + x = c`

Answer 1

The general solution of `(1 + e^{x/y})dx + e^{x/y}(1 - x/y)dy = 0` is `underline(ye^{x/y} + x = c)`.

Explanation:

`(1 + e^{x/y})dx + e^{x/y}(1 - x/y)dy = 0`

⇒ `dx/dy = (-e^{x/y}(1 - x/y))/(1 + e^{x/y})` ............(i)

Put `x/y = u` ............(ii)

⇒ `dx/dy = u + y(du)/dy` ............(iii)

Substituting (ii) and (iii) in (i), we get

`u + y (du)/dy = (-e^u(1 - u))/(1 + e^u)`

∴ `y (du)/dy = (-e^u - u)/(1 + e^u)`

∴ `(1 + e^u)/(u + e^u) du = (-dy)/y`

Integrating on both sides, we get

`int(1 + e^u)/(u + e^u) du = -intdy/y`

∴ log|u + e^u| = -log|y| + log|c|

∴ `log|x/y + e^{x/y}| = log|c/y|`

∴ `x/y + e^{x/y} = c/y`

∴ `ye^{x/y} + x = c`