ODE with power series and non-zero initial conditions

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I should resolve
$$y''-(1+x)y=0$$
First for $y(0)=-3,y'(0)=6$. In this case, I got some values until for $a_5$ but I didn't see anything between the $a_n$, so I got:
$$y(x)=frac212+sum_n=3^inftyleft[(n+1)(n+2)a_n+2+a_n-1right]x^n$$



I don't know if it's right but I have a bigger problem: what am I supposed to do when the initial conditions are $y(2)=4$ and $y'(2)=3$?







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  • 4




    Being very lazy, I suppose that I should define $t=x-2$ and make the series in terms of $t$.
    – Claude Leibovici
    Jul 16 at 4:12














up vote
0
down vote

favorite












I should resolve
$$y''-(1+x)y=0$$
First for $y(0)=-3,y'(0)=6$. In this case, I got some values until for $a_5$ but I didn't see anything between the $a_n$, so I got:
$$y(x)=frac212+sum_n=3^inftyleft[(n+1)(n+2)a_n+2+a_n-1right]x^n$$



I don't know if it's right but I have a bigger problem: what am I supposed to do when the initial conditions are $y(2)=4$ and $y'(2)=3$?







share|cite|improve this question















  • 4




    Being very lazy, I suppose that I should define $t=x-2$ and make the series in terms of $t$.
    – Claude Leibovici
    Jul 16 at 4:12












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I should resolve
$$y''-(1+x)y=0$$
First for $y(0)=-3,y'(0)=6$. In this case, I got some values until for $a_5$ but I didn't see anything between the $a_n$, so I got:
$$y(x)=frac212+sum_n=3^inftyleft[(n+1)(n+2)a_n+2+a_n-1right]x^n$$



I don't know if it's right but I have a bigger problem: what am I supposed to do when the initial conditions are $y(2)=4$ and $y'(2)=3$?







share|cite|improve this question











I should resolve
$$y''-(1+x)y=0$$
First for $y(0)=-3,y'(0)=6$. In this case, I got some values until for $a_5$ but I didn't see anything between the $a_n$, so I got:
$$y(x)=frac212+sum_n=3^inftyleft[(n+1)(n+2)a_n+2+a_n-1right]x^n$$



I don't know if it's right but I have a bigger problem: what am I supposed to do when the initial conditions are $y(2)=4$ and $y'(2)=3$?









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 16 at 2:31









mvfs314

422210




422210







  • 4




    Being very lazy, I suppose that I should define $t=x-2$ and make the series in terms of $t$.
    – Claude Leibovici
    Jul 16 at 4:12












  • 4




    Being very lazy, I suppose that I should define $t=x-2$ and make the series in terms of $t$.
    – Claude Leibovici
    Jul 16 at 4:12







4




4




Being very lazy, I suppose that I should define $t=x-2$ and make the series in terms of $t$.
– Claude Leibovici
Jul 16 at 4:12




Being very lazy, I suppose that I should define $t=x-2$ and make the series in terms of $t$.
– Claude Leibovici
Jul 16 at 4:12















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