Question

Solve (x + 1)dy/dx-y = e^3x(x + 1)^3

Answer 1

Given :

(x + 1) dy/dx - y = e³ˣ (x + 1)³

Solution :

Divide both sides by (x + 1)

dy/dx - y/(x + 1) = e³ˣ (x + 1)²

dy/dx + Py = Q

So, P = -1/1+x

Q = e³ˣ (x + 1)²

$IF = e^{\int\limits{P} \, dx }$

$IF = e^{\int\limits{-1/1+x} \, dx }$

$IF = e^{-log(1+x)}$

IF = 1 / (1 + x)

General solution :

$y*IF = \int\limits {Q*IF} \, dx$

$y*1/(1+x) = \int\limits {e^{3x}*(x+1)^{2}*(1/(1+x)) } \, dx$

$y*1/(1+x) = \int\limits {e^{3x}*(x+1) } \, dx$

$y / (1 +x) =(x+1)\int\limits {e^{3x} } \, dx -\int\limits( {d(x+1)/dx*\int\limits{e^{3x} } \, dx }) \, dx$

$y/(x+1) = (x+1)e^{3x} /3 - \int\limits {e^{3x} / 3} \, dx$

$y/(1+x) = (x+1)e^{3x}/3 - e^{3x} /9$

$y=(x+1)^{2} e^{3x}/3 - (x+1)e^{3x}/9 + c$