# Solve second order diff eq with substitution

1. Jun 28, 2011

### Asphyxiated

1. The problem statement, all variables and given/known data

Solve:

$$xy''+2y'=12x^{2}$$

with

$$u=y'$$

2. Relevant equations

if you have:

$$y'+P(x)y=Q(x)$$

then your integrating factor is:

$$I(x)=e^{\int P(x) dx}$$

3. 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:

$$xy''+2y'=12x^{2} \hookrightarrow y''+\frac{2y'}{x}=12x$$

and

$$y'=u \hookrightarrow y''=u\frac{du}{dy}$$

I do not understand why y'' is equal to this. The way I read y'' is d2y/dx2 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:

$$u\frac{du}{dy}+\frac{2u}{x}=12x$$

then

$$I(x)=e^{\int \frac {2}{x} dx} = e^{2 ln|x|}=x^{2}$$

then multiply both sides by I(x):

$$x^{2}u\frac{du}{dy}+2xu=12x^{3}$$

which is:

$$\frac {d}{dx} (ux^{2})=12x^{3}$$

$$ux^{2}=\int 12x^{3} dx =3x^{4}+C_{1}$$

$$u=3x^{2}+ \frac{C_{1}}{x^{2}}$$

$$\frac {dy}{dx}=3x^{2} + \frac {C_{1}}{x^{2}}$$

$$y= \int 3x^{2}+\frac{C_{1}}{x^{2}}dx = x^{3}-\frac{C_{1}}{x}+C_{2}$$

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

2. Jun 28, 2011

### vela

Staff Emeritus
That's just the chain rule.
$$y'' = \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx} = u\frac{du}{dy}$$
I'm not sure why you'd do it that way, though. You could simply write
$$x u' + 2 u = 12x^2$$
and solve it with the same integrating factor without the unnecessary complication of invoking the chain rule.

3. Jun 28, 2011

### Asphyxiated

oh haha alright. I guess I was confused by the substitution of y'=u again for some reason.
Thanks for the reminder on that!

So I assume this is a correct solution? It's an even problem so I can't look it up.

4. Jun 28, 2011

### vela

Staff Emeritus
I didn't see anything obviously wrong, but I didn't look that closely either. You can always check your answer by plugging it back into the original differential equation to see if it's satisfied.