Answer to Question #251705 in Differential Equations for jor

Question #251705

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


1
Expert's answer
2021-10-17T17:44:38-0400
(3xy+3y4)dx+(x+1)2dy=0(3xy+3y-4)dx+(x+1)^2dy=0

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

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

Integrating factor


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

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

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

Integrate


d((x+1)3y)=4(x+1)dx\int d((x+1)^3y)=\int4(x+1)dx

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

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


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