# Second order diff.eq. help?

1. Feb 14, 2008

### sutupidmath

Second order diff.eq. help??

well i am trying to find a solution to this diff. eq, but i get stuck somewhere.
$$\ 4x^{2}y''+y=0$$

I first took this substitution

y'=p, y"=p' so the diff. eq becomes of this form

$$\ 4x^{2}p'+p=0$$
i think this can be done with the separable of variables thing. so
$$\ 4x^{2}\frac{dp}{dx}=-p$$

$$\frac{dp}{p}=-\frac{dx}{4x^{2}}$$ now i integrate bot sides
$$\int\frac{dp}{p}$$ = $$-\int\frac{dx}{4x^{2}}$$, after some calculations i get

$$p=A e^{\frac{1}{4x}$$, where A is a constant that we get because of integrating bot siedes. NOw i go back to p=y'
so

$$y'= A e^{\frac{1}{4x}$$, now i think all i need to do is integrate, but i get stuck integrating the right part of this eq. I do not think it has any closed form, right?

but i know that the answer to this equation is in a closed form, that is, it is an elementary function, so where am i going wrong?
I also tried to consider the eq $$\ 4x^{2}p'+p=0$$ as a linear one, but also when i tried to derive an integrating factor i ended up with a similar expression as $$y'= A e^{\frac{1}{4x}$$,
so, any hints on how to solve this?

2. Feb 14, 2008

### Rainbow Child

You can not apply the substitution $y'(x)=p(x),\,y''=p'(x)$ in your equation since it does not contains $y'(x)$
Try the substitution $y(x)=\sqrt{x}\,p(x)$ to arrive to $x\,p''(x)+p'(x)=0$ and then try your substitution. i.e. $p'(x)=w(x),\,p''(x)=w'(x)$

3. Feb 14, 2008

### sutupidmath

How does one know that such a substitution will work?? Is there a theorem that provides us with these kind of substitutions or???

4. Feb 14, 2008

### John Creighto

I'm not sure how the above works. Why not substitute into the equation y=x^n and then solve for n. I say this because if you differentiate twice you lose two powers of x.

5. Mar 10, 2008

### kthouz

Why can't you use the Power Series Method where you let the solution y=$$\sum_{0}a_n{}x^{n}$$ where a is from to infinite. The you will replace y along with its derivatives in that equation

6. Mar 10, 2008

### sutupidmath

I guess i have to wait for another month then, untill we get to that chapter! We haven't yet done this method of inifinite taylor series, or whatever!.

7. Mar 10, 2008

### Peeter

DEs with this form are called Euler differential equations. A substuition of the form:

z = ln(x)

with $$\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx}$$, can be used to convert this to a constant coefficient second order equation.