Question

Find the general solution of (2x+1)y"-2y'-(2x+3)y=(2x+1)^2, given that y1=e^-x satisfies the complementary equation.

Answer 1

y=ue^-x
y'=u'e^(-x)-ue^(-x)
y"=u"e^(-x)-2u'e^(-x)+ue^(-x)
Subbing:
(2x+1)y"-2y'-(2x+3)y=(2x+1)(u"e^(-x)-2u'e^(-x)+ue^(-x))-2(u'e^(-x)-ue^(-x))-(2x+3)ue^(-x)=u"2xe^(-x)+u"e^(-x)-4u'xe^(-x)-4u'e^(-x)=e^(-x)(u"2x+u"-4u'x-4u')

So,
e^(-x)(u"2x+u"-4u'x-4u')=(2x+1)^2
u"2x+u"-4u'x-4u'=e^(x)(2x+1)^2
Am I right up to this point?
@wio

Answer 2

Hard to read

Answer 3

I'm not sure, actually. I haven't done this in a while

Answer 4

@ganeshie8

Answer 5

I think you're correct up to that step

Answer 6

Okay, so what should I do next then?

Answer 7

What other methods do you know?

Answer 8

They don't teach you want to do when you have variable coefficients?

Answer 9

No, it's because at the end, I got some weird results, that's why. If you said up to that point I'm right, then this is what I got next from above.

u"(2x+1)-4u'(x+1)=e^(x)(2x+1)^2
\[u''-\frac{ 4u'(x+1) }{ 2x+1 }=e ^{x}(2x+1)\]

Answer 10

Then z=u', subbing:
\[z'-\frac{ 4(x+1) }{ 2x+1 }z=e ^{x}(2x+1)\]

Answer 11

Up to this point above, I tried to integrate the middle term because I applied the method of integrating factor in order to find z at the end and integrate u in order to find y.

Answer 12

Any other method to solve for z without doing the integrating factor method?

Answer 13

Seems like you should just bite the bullet and do your partial fractions