Answer to Question #251705 in Differential Equations for jor

Question

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

Answer 1

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

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

$$y'+\dfrac{3}{x+1}y=\dfrac{4}{(x+1)^2}$$

Integrating factor

$$\mu(x)=e^{\int({3 \over x+1})dx}=(x+1)^3$$

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

$$d((x+1)^3y)=4(x+1)dx$$

Integrate

$$\int d((x+1)^3y)=\int4(x+1)dx$$

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

$$y=\dfrac{2}{x+1}+\dfrac{C}{(x+1)^3}$$