How can Euler differential equations be solved using a substitution?

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sutupidmath
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Second order diff.eq. help??

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

I first took this substitution

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

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

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

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

[tex]y'= A e^{\frac{1}{4x}[/tex], 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 [tex]\ 4x^{2}p'+p=0[/tex] as a linear one, but also when i tried to derive an integrating factor i ended up with a similar expression as [tex]y'= A e^{\frac{1}{4x}[/tex],
so, any hints on how to solve this?
 
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You can not apply the substitution [itex]y'(x)=p(x),\,y''=p'(x)[/itex] in your equation since it does not contains [itex]y'(x)[/itex]
Try the substitution [itex]y(x)=\sqrt{x}\,p(x)[/itex] to arrive to [itex]x\,p''(x)+p'(x)=0[/itex] and then try your substitution. i.e. [itex]p'(x)=w(x),\,p''(x)=w'(x)[/itex]
 
Rainbow Child said:
Try the substitution [itex]y(x)=\sqrt{x}\,p(x)[/itex] to arrive to [itex]x\,p''(x)+p'(x)=0[/itex]
How does one know that such a substitution will work?? Is there a theorem that provides us with these kind of substitutions or?
 
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.
 
Why can't you use the Power Series Method where you let the solution y=[tex]\sum_{0}a_n{}x^{n}[/tex] where a is from to infinite. The you will replace y along with its derivatives in that equation
 
kthouz said:
Why can't you use the Power Series Method where you let the solution y=[tex]\sum_{0}a_n{}x^{n}[/tex] where a is from to infinite. The you will replace y along with its derivatives in that equation

I guess i have to wait for another month then, until we get to that chapter! We haven't yet done this method of inifinite taylor series, or whatever!.
 
sutupidmath said:
well i am trying to find a solution to this diff. eq, but i get stuck somewhere.
[tex]\ 4x^{2}y''+y=0[/tex]

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

z = ln(x)

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