A equation -series solution

[georgelymp]

Solve the equation y'' + (x^2)y' +2xy=0 ,x0=0.

[Guest]

Since you are told what to do- look for a series solution- why haven't you at least tried doing that?

A "series solution" is something like [tex]y= \sum_{n=0}^\infty a_nx^n[/tex]. Then [tex]y'= \sum_{n=1}^\infty na_nx^{n-1}[/tex] and [tex]y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}[/tex]. Putting those into [tex]y''+ x^2y'+ 2xy= 0[/tex] gives [tex]\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=1}^\infty na_nx^{n+1}+ \sum_{n=0}^\infty 2a_nx^{n+1}= 0[/tex]. Get the same power of x in each term by letting i= n- 2, so that n= i+2, in the first sum and letting i= n+1, so that n= i-1 in the other two. That is, with i= n-2, [tex]y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}= \sum_{i=0}^\infty (i+2)(i+1)a_{i+1}x^i[/tex]. With i=n+1, [tex]x^2y'= \sum_{n=1}^\infty a_nx^{n+1}= \sum_{i=2}^\infty (i+1)a_{i-1}x^i[/tex]. With i= n+1, [tex]2xy= \sum_{n=0}^\infty 2a_nx^{n+1}= \sum_{i=1}^\infty 2a_{i-1}a_{i-1}x^i[/tex].

That gives [tex]\sum_{i=0}^\infty (i+2)(i+1)a_{i+2}x^i+ \sum_{i=2}^\infty (i-1)a_{i-1}x^i+ \sum_{i=1}^\infty 2a_{i-1}x^i= 0[/tex]. In order for that to be 0 for all x, the coefficient of each power of x must be 0. When i= 0 that give [tex]2a_2= 0[/tex], when i= 1, [tex]6a_3+ 2a_0= 0[/tex], and when [tex]i\ge 2[/tex], [tex](i+2)(i+1)a_{i+2}+ (i-1)a_{i-1}+2a_{i-1}= 0[/tex]. Notice that neither [tex]a_0[/tex] nor [tex]a_1[/tex] is determined. They serve as the "undetermined constants" which will be determined by the initial conditions.