Yay, another differential equations question

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SUMMARY

The discussion focuses on solving the differential equation y'' + p(x)y' + q(x)y = r(x) using a particular solution method. The participants demonstrate that if y1(x) is a solution to the homogeneous equation, then y2 = u(x)y1(x) is a solution to the non-homogeneous equation, provided that v = u' is a solution to the associated linear differential equation. The conversation emphasizes the importance of integrating factors, specifically denoted as w, to simplify the solution process and manage the resulting integrals effectively.

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schattenjaeger
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if you have a DE y''+p(x)y'+q(x)y=r(x) where p, q, and r are continuous on interval I, if y=y1(x) is a solution to the associated homogenous equation, show that y2=u(x)y1(x) is a solution to the DE provided v=u' is a solution to the linear DE.
v'+v[2(y'1/y1)+p]=r/y1
Express the solution to the DE in terms of integrals and so on...

well ok then, I started with y2=u*y1 so y'2=u'*y1+u*y'1 and y''2=u''y1+2u'y'1+uy''1

I plugged those back into the original equation, and got u*(y1''+py'1+qy1) + u''y1+u'(2y'1+py1)=r

knowing that first part was just 0 since y1 is a solution to the homogenous equation, I can rearrange everything to

u''+u'[2(y'1/y1)+p]=r/y1

sure enough that's the other equation I was given where I was told v=u' is a solution. Question: what good does that do me and where do I go from there? I mean, I could then just sub u with v, solve for v, take the integral to get u, and multiply by y1 but then, in between all that integrating, I need an integrating factor, and I end up with like integrals in the exponentials and integrals all over the place, to the point where I know I slipped up...
 
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did I mention I'll love you forrreverrr?
 
This is easy. Be glad you need only write integral, you need not compute, any integrals. You should post your integrals in the exponentials and integrals all over the place, that sound like fun. Anyway you used u and v, thus taking away my joy in using those letters to represent integrating factors. The next best thing is to call the integrating factor w.
Thus it is desired to find w such that
(w*u')'=w*u''+w*u'[2(y1'/y1)+p]
w*u''+u'w'=w*u''+w*u'[2(y1'/y1)+p]
you should be able to manage that
next you would multiply
u''+u'[2(y1'/y1)+p]=r/y1
by w
w*u''+w*u'[2(y1'/y1)+p]=w*r/y1
note w is an integrating factor
(w*u')'=w*r/y1
so w is some integral
u can be found with two more integrals
there will indeed be some exponentials of integrals
or as people say big fun
your final answer will look neater if you find w and write out the integral, then use w in further work as a symbol rather that writing out the integral where ever w appears.
 

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