# 2nd order differential eguation

1. Apr 8, 2013

### MostlyHarmless

1. The problem statement, all variables and given/known data
Here is the problem, verbatim.

Observe that y=x is a particular solution of the equation $$2x^2y''+xy'-y=0[\tex] and find the general solution. For what values of x is the solution valid? 2. Relevant equations 3. The attempt at a solution I know the answer is $y=c_1x+c_2x^{{\frac{-1}{2}}}$, but I cannot, for the life of me, figure out how on Earth I'm even supposed to go about finding that solution. It should be noted that this is in the section for The Reduction of Order Method, so I assume it has to with that, however, everything we've done thus far, since the introduction of differential operators, has only concerned Diff. Eq. w/ constant coefficients. So thats where I'm hitting my main roadblock. I've tried several things, none of which got me anywhere. For example: I tried re-writing the original expression as [tex](2x^2D^2+xD-1)y=0$$ and letting $y=c_1x+c_2y_1$ I quickly realized that just takes me back to my original expression.

I then tried letting y=vx, as per reduction of order method suggests. That would give $y'=v'x+v$ and $y''=v''x+2v'$
plugging that into the original equation:

$2x^3v''+4x^2v'+x^2v'+xv-xv=0$

The xv's will subtract out. Leaving

$2x^3v''+5x^2v'=0$

dividing by $2x^3$ gives
$$v''+{\frac{5}{2}}x^{-1}v'=0$$
Then I made the substitution w=v'; w'=v'' to make it first order linear in w.
$$w'+{\frac{5}{2}}x^{-1}w=0$$
giving an integrating factor of

$e^{∫{\frac{5}{2}}x^{-1}}=x^{\frac{5}{2}}$ Multiplying through I get $$d[wx^{\frac{5}{2}}]=0$$ Then thats where I get stuck. Integrating now will eventually give v=0, which is unhelpful.

2. Apr 8, 2013

### Dick

That's a Euler equation. You can tell by the way the powers of x match up with the order of the differentiation. The appropriate substitution to try is $y=x^n$. Try to solve for n.

3. Apr 8, 2013

### LCKurtz

Don't give up, you are almost there. $(wx^{\frac 5 2})' = 0$ gives $wx^{\frac 5 2}=C$ or $w = Cx^{-\frac 5 2}$. So what is $v$? Then what do you get for your second solution $xv$? And you don't need to worry about the multiplying constant since any constant times a solution is a solution.

4. Apr 8, 2013

### LCKurtz

Don't you think that misses the point of the question? They are learning reduction of order.

5. Apr 8, 2013

### Dick

Yes, I did gloss over that point. I just picked the easiest way to a solution. Apologies.

6. Apr 8, 2013

### MostlyHarmless

So just do the same method I'm doing, but rather than y=vx use $y=x^n$? Or is there a specific way of solving a Euler equation? I've not studied them before, but I can find out how to do them fairly quickly if you wouldnt mind giving me one more nudge.

7. Apr 8, 2013

### MostlyHarmless

Oh ok, I can't believe I forgot my constant!

8. Apr 8, 2013

### MostlyHarmless

Alright, I got: $∫v'=∫c_1x^{\frac{-5}{2}}dx$ Which gives $v=c_1x^{\frac{-3}{2}}+c_2$ multiplying through by x to get back to y gives: $y=c_1x^{\frac{-1}{2}}+c_2x$ which was the correct answer.

Thank you guys. I'd been struggling with that for awhile, I was just forgetting the constant!!

9. Apr 8, 2013

### Dick

And recognizing it as a Euler equation does give you a huge shortcut which doesn't use reduction of order. Put y=x^n cancel the x^n's and just solve for n. Try it. It's a pretty good trick to know.

Last edited: Apr 8, 2013