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## Homework Statement

Solve:

[tex] xy''+2y'=12x^{2} [/tex]

with

[tex] u=y' [/tex]

## Homework Equations

if you have:

[tex] y'+P(x)y=Q(x) [/tex]

then your integrating factor is:

[tex] I(x)=e^{\int P(x) dx} [/tex]

## The Attempt at a Solution

The only reason I was able to solve this is because I stumbled upon a similar post here but instead of pulling up a thread from over a year ago i decided to post my specific question.

So if:

[tex] xy''+2y'=12x^{2} \hookrightarrow y''+\frac{2y'}{x}=12x[/tex]

and

[tex] y'=u \hookrightarrow y''=u\frac{du}{dy} [/tex]

I do not understand why y'' is equal to this. The way I read y'' is d

^{2}y/dx

^{2}so why is it u*du/dy as the derivative on the right side and not d/dx? And! why is the derivative of u with respect to y u*du/dy and not just du/dy.

Maybe I have just forgotten something simple from differential calculus but I can't make sense of it on my own. I am not claiming that what is written in TeX above is wrong because the problem seems to work out, I just don't know why that is the way to do it.

Anyway, if we make the substitution:

[tex] u\frac{du}{dy}+\frac{2u}{x}=12x [/tex]

then

[tex] I(x)=e^{\int \frac {2}{x} dx} = e^{2 ln|x|}=x^{2} [/tex]

then multiply both sides by I(x):

[tex] x^{2}u\frac{du}{dy}+2xu=12x^{3} [/tex]

which is:

[tex] \frac {d}{dx} (ux^{2})=12x^{3} [/tex]

[tex] ux^{2}=\int 12x^{3} dx =3x^{4}+C_{1} [/tex]

[tex] u=3x^{2}+ \frac{C_{1}}{x^{2}} [/tex]

[tex] \frac {dy}{dx}=3x^{2} + \frac {C_{1}}{x^{2}} [/tex]

[tex] y= \int 3x^{2}+\frac{C_{1}}{x^{2}}dx = x^{3}-\frac{C_{1}}{x}+C_{2} [/tex]

if that is right please tell me why. I just followed by example from someone elses work.