Question

Solve the differential equation `(dy)/(dx) = (4x + y + 1)^2`.

Answer 1

`(dy)/(dx) = (4x + y + 1)^2`   ...(1)

Put 4x + y + 1 = u

∴ `4 + (dy)/(dx) = (du)/(dx)`

∴ `(dy)/(dx) = (du)/(dx) - 4`

∴ (1) becomes, `(du)/(dx) - 4 = u^2`

∴ `(du)/(dx) = u^2 + 4`

∴ `1/(u^2 + 4) du = dx`

Integrating both sides, we get

`int 1/(u^2 + 2^2) du = int dx`

∴ `1/2 tan^-1 (u/2) = x + c_1`

∴ `tan^-1 ((4x + y + 1)/(2)) = 2x + 2c_1`

∴ `tan^-1 ((4x + y + 1)/2) = 2x + c`, where `c = 2c_1`

This is the general solution.