Question

The solution of the differential equation `(dy)/(dx) = 1 + x + y + xy` when y = 0 at x = – 1 is

Options

• `e^(1/2(1 + x^2)`

• `e^(1/2(1 + x^2)) - 1`

• `e^(1/2(1 + x^2)) + 1`

• None of these

Answer 1

None of these

Explanation:

Given `(dy)/(dx) = 1 + x + y + xy = (1 + x) + y(1 + x)`

`(dy)/(dx) = (1 + x)(1 + y)`

`int 1/(1 + y)  dy = int (1 + x)  dx`

`log |1 + y| = x + x^2/2 + c`

Put `y = 0, x = - 1`

`log 1 = - 1 + 1/2 + c` ⇒ `c = 1/2`

∴ `log|1 + y| = x + x^2/2 + 1/2`

`y = e^(2x) (+ x^2 + 1)/2 - 1`

`y = e^(1/2)(1 + x)^2 - 1`.