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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...

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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