- #1
sutupidmath
- 1,630
- 4
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?
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?