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use power series method to find solution of $(x^2+2x)y''+(x^2-2)y'-(2x+2)y=0$ i have tried $y=\sum_{0}^{\infty}{am*x^{m+r}}$ normalize the power to $m+r-1$ ,for m=0, $(2r^2-4r)a_0=0$, $r=0,2$ or a0=0 , m=1 , $(r^2-r+2)a_0=0$

m≥2

$2(m+r)(m+r-2)a_m+[(m+r-1)(m+r-2)+2]a_m-1+(m+r-4)a_m-2=0$

then i don't know the next step to find the regular pattern, can anyone help me?

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I have no idea how you get the identity just after "$m\ge2$". Can you explain? –  Did Dec 6 '13 at 7:11
    
sorry,i have edited the formula again. –  user113644 Dec 6 '13 at 8:09

1 Answer 1

The approach suggested to you, only implemented more rigorously, is the following. Assume that $y(x)=\sum\limits_na_nx^{n+r}$ solves the ODE, then $$y'(x)=\sum\limits_n(n+r+1)a_{n+1}x^{n+r},\quad y''(x)=\sum\limits_n(n+r+2)(n+r+1)a_{n+2}x^{n+r}, $$ and, for every integer $i$, $$ x^iy(x)=\sum\limits_na_{n-i}x^{n+r},\quad x^iy'(x)=\sum\limits_n(n+r+1-i)a_{n+1-i}x^{n+r}, $$ and $$ x^iy''(x)=\sum\limits_n(n+r+2-i)(n+r+1-i)a_{n+2-i}x^{n+r}. $$ Hence, the first term of the ODE becomes $$ (x^2+2x)y''(x)=\sum\limits_n(n+r)(n+r-1)a_{n}+2(n+r+1)(n+r)a_{n+1} x^{n+r}, $$ and similarly for the other terms. Putting everything together yields a relation between $a_n$, $a_{n+1}$ and $a_{n-1}$, to be solved. The initial condition comes from the terms of lowest degree. These are the initial term of $2xy''-2y'$, thus $$ 2r(r-1)a_0-2ra_0=0, $$ which yields $r=0$ or $r=2$, hence everything is in place to solve the ODE.


Another approach, more direct, is to note that, considering $a(x)=x^2+2x$, the ODE becomes $$ a(x)y''(x)+(a(x)-a'(x))y'(x)-a'(x)y(x)=0, $$ that is, considering $z(x)=y'(x)+y(x)$, $$ a(x)z'(x)-a'(x)z(x)=0. $$ Thus, there exists some constant $\lambda$ such that $$z(x)=\lambda a(x). $$ Now, $a(x)=b(x)+b'(x)$ where $b:x\mapsto x^2$ hence there exists some $\lambda$ such that $$ y'(x)+y(x)=\lambda (b(x)+b'(x)), $$ that is, $$ (y-\lambda b)'(x)=-(y-\lambda b)(x), $$ hence there exists some $\mu$ such that $$ (y-\lambda b)(x)=\mu\mathrm e^{-x}, $$ that is, $$ y(x)=\lambda x^2+\mu\mathrm e^{-x}. $$ Note that the case $r=0$ in the first approach corresponds to $\mu\ne0$ while the case $r=2$ corresponds to $\mu=0$ and $\lambda\ne0$.

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thank you so much!!the second method is so amazing! –  user113644 Dec 7 '13 at 1:59

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