Answer to Question #144875 in Differential Equations for Divine Grace Jaravata

Question

( 3xy + 3y - 4)dx + (x + 1)2dy = 0

Answer 1

Let us solve the equation $( 3xy + 3y - 4)dx + (x + 1)^2dy = 0$:

$3xy + 3y - 4 + (x + 1)^2y' = 0$

$3(x + 1)y + (x + 1)^2y' = 4$

Multiply both parts of the differential equation by $x+1:$

$3(x + 1)^2y + (x + 1)^3y' = 4(x+1)$

$((x + 1)^3y)' = 4x+4$

Therefore, we have the following solution:

$(x + 1)^3y = 2x^2+4x+C$