Yay, another differential equations question

AI Thread Summary
The discussion focuses on solving a differential equation of the form y'' + p(x)y' + q(x)y = r(x) using a known solution y1(x) of the associated homogeneous equation. It explores the substitution y2 = u(x)y1(x) and derives a new equation for u, leading to the expression u'' + u'[2(y1'/y1) + p] = r/y1. Participants discuss the challenges of integrating and finding an integrating factor, suggesting the use of a new variable w for clarity. The conversation emphasizes that while the process involves complex integrals, the final solution can be simplified by properly defining w and expressing integrals neatly. Overall, the thread highlights the intricacies of solving differential equations and the importance of organization in mathematical expressions.
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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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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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