How simplify final answer for series solution for $x''(t) + 2tx'(t) − 8x(t) = 0$?











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We are given the IVP $x''(t) + 2tx'(t) − 8x(t) = 0$ subject to $x(0)=1$ and $x'(0)=0$.
I got:



$x(t)$ = $sum_{n=0}^∞=a_n.t^n$



$x'(t)= sum_{n=0}^∞=n.a_n.t^{n-1}$



$x''(t)= sum_{n=0}^∞=(n+1)(n+2).a_{n+2}.t^{n}$ using shift of sumbation.



Subbing in i got and simplifying i got:



$sum_{n=0}^∞=[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]t^n$.



And we want $[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]=0$



$a_{n+2} = (8-2n).a_n/(n+2)(n+1)$



We have $x(0)=1$ and $x'(0)=0$, so $a_o= 1$ and $a_1 = 0$, and we can get the rest of the solutions:



$a_2=4,a_3=0, a_4=4/3,a_5=0,a_6=0,a_7=0,a_8=0,a_9=0$



Have i attempted this correctly, and how do i get the final answers? Im not seeing the connection between the solutions..






sequences-and-series differential-equations







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  • 2




    You get $a_k=0$ for $k>4$, so the solution is a polynomial of degree 4, $x(t)=1+4x^2+frac43x^4$, as per your computed coefficients.
    – LutzL
    Nov 22 at 21:13















up vote
1
down vote

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We are given the IVP $x''(t) + 2tx'(t) − 8x(t) = 0$ subject to $x(0)=1$ and $x'(0)=0$.
I got:



$x(t)$ = $sum_{n=0}^∞=a_n.t^n$



$x'(t)= sum_{n=0}^∞=n.a_n.t^{n-1}$



$x''(t)= sum_{n=0}^∞=(n+1)(n+2).a_{n+2}.t^{n}$ using shift of sumbation.



Subbing in i got and simplifying i got:



$sum_{n=0}^∞=[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]t^n$.



And we want $[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]=0$



$a_{n+2} = (8-2n).a_n/(n+2)(n+1)$



We have $x(0)=1$ and $x'(0)=0$, so $a_o= 1$ and $a_1 = 0$, and we can get the rest of the solutions:



$a_2=4,a_3=0, a_4=4/3,a_5=0,a_6=0,a_7=0,a_8=0,a_9=0$



Have i attempted this correctly, and how do i get the final answers? Im not seeing the connection between the solutions..






sequences-and-series differential-equations







share|cite|improve this question




















  • 2




    You get $a_k=0$ for $k>4$, so the solution is a polynomial of degree 4, $x(t)=1+4x^2+frac43x^4$, as per your computed coefficients.
    – LutzL
    Nov 22 at 21:13













up vote
1
down vote

favorite









up vote
1
down vote

favorite











We are given the IVP $x''(t) + 2tx'(t) − 8x(t) = 0$ subject to $x(0)=1$ and $x'(0)=0$.
I got:



$x(t)$ = $sum_{n=0}^∞=a_n.t^n$



$x'(t)= sum_{n=0}^∞=n.a_n.t^{n-1}$



$x''(t)= sum_{n=0}^∞=(n+1)(n+2).a_{n+2}.t^{n}$ using shift of sumbation.



Subbing in i got and simplifying i got:



$sum_{n=0}^∞=[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]t^n$.



And we want $[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]=0$



$a_{n+2} = (8-2n).a_n/(n+2)(n+1)$



We have $x(0)=1$ and $x'(0)=0$, so $a_o= 1$ and $a_1 = 0$, and we can get the rest of the solutions:



$a_2=4,a_3=0, a_4=4/3,a_5=0,a_6=0,a_7=0,a_8=0,a_9=0$



Have i attempted this correctly, and how do i get the final answers? Im not seeing the connection between the solutions..






sequences-and-series differential-equations







share|cite|improve this question















We are given the IVP $x''(t) + 2tx'(t) − 8x(t) = 0$ subject to $x(0)=1$ and $x'(0)=0$.
I got:



$x(t)$ = $sum_{n=0}^∞=a_n.t^n$



$x'(t)= sum_{n=0}^∞=n.a_n.t^{n-1}$



$x''(t)= sum_{n=0}^∞=(n+1)(n+2).a_{n+2}.t^{n}$ using shift of sumbation.



Subbing in i got and simplifying i got:



$sum_{n=0}^∞=[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]t^n$.



And we want $[(n+2)(n+1)a_{n+2} +2n.a_n-8a_n]=0$



$a_{n+2} = (8-2n).a_n/(n+2)(n+1)$



We have $x(0)=1$ and $x'(0)=0$, so $a_o= 1$ and $a_1 = 0$, and we can get the rest of the solutions:



$a_2=4,a_3=0, a_4=4/3,a_5=0,a_6=0,a_7=0,a_8=0,a_9=0$



Have i attempted this correctly, and how do i get the final answers? Im not seeing the connection between the solutions..






sequences-and-series differential-equations




sequences-and-series differential-equations






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edited Nov 22 at 20:09

























asked Nov 22 at 19:57









Neels

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  • 2




    You get $a_k=0$ for $k>4$, so the solution is a polynomial of degree 4, $x(t)=1+4x^2+frac43x^4$, as per your computed coefficients.
    – LutzL
    Nov 22 at 21:13














  • 2




    You get $a_k=0$ for $k>4$, so the solution is a polynomial of degree 4, $x(t)=1+4x^2+frac43x^4$, as per your computed coefficients.
    – LutzL
    Nov 22 at 21:13








2




2




You get $a_k=0$ for $k>4$, so the solution is a polynomial of degree 4, $x(t)=1+4x^2+frac43x^4$, as per your computed coefficients.
– LutzL
Nov 22 at 21:13




You get $a_k=0$ for $k>4$, so the solution is a polynomial of degree 4, $x(t)=1+4x^2+frac43x^4$, as per your computed coefficients.
– LutzL
Nov 22 at 21:13










1 Answer
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The recurrence equation can actually be solved. The coefficient of $t^{2k+1}$ turns out to be $$ frac{3(-1)^k}{k!(2k-3)(4k^2-1)}$$
It turns out that the solution can be expressed in closed form:



$$ x(t) = frac{2t^3+5t}{8} e^{-t^2} + frac{sqrt{pi} text{erf}(t) (4 t^4 + 12 t^2 + 3)}{16}$$



although getting that from the series is not going to be easy.



EDIT:
Too bad you changed the initial conditions. Now the solution is as given by LutzL. The original was much more interesting, I think.






share|cite|improve this answer























  • Thanks for the reply. How do you figure this out so quickly?
    – Neels
    Nov 22 at 20:12










  • Also please check my edits, because i initially typed out the wrong limits.
    – Neels
    Nov 22 at 20:13











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up vote
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The recurrence equation can actually be solved. The coefficient of $t^{2k+1}$ turns out to be $$ frac{3(-1)^k}{k!(2k-3)(4k^2-1)}$$
It turns out that the solution can be expressed in closed form:



$$ x(t) = frac{2t^3+5t}{8} e^{-t^2} + frac{sqrt{pi} text{erf}(t) (4 t^4 + 12 t^2 + 3)}{16}$$



although getting that from the series is not going to be easy.



EDIT:
Too bad you changed the initial conditions. Now the solution is as given by LutzL. The original was much more interesting, I think.






share|cite|improve this answer























  • Thanks for the reply. How do you figure this out so quickly?
    – Neels
    Nov 22 at 20:12










  • Also please check my edits, because i initially typed out the wrong limits.
    – Neels
    Nov 22 at 20:13















up vote
0
down vote













The recurrence equation can actually be solved. The coefficient of $t^{2k+1}$ turns out to be $$ frac{3(-1)^k}{k!(2k-3)(4k^2-1)}$$
It turns out that the solution can be expressed in closed form:



$$ x(t) = frac{2t^3+5t}{8} e^{-t^2} + frac{sqrt{pi} text{erf}(t) (4 t^4 + 12 t^2 + 3)}{16}$$



although getting that from the series is not going to be easy.



EDIT:
Too bad you changed the initial conditions. Now the solution is as given by LutzL. The original was much more interesting, I think.






share|cite|improve this answer























  • Thanks for the reply. How do you figure this out so quickly?
    – Neels
    Nov 22 at 20:12










  • Also please check my edits, because i initially typed out the wrong limits.
    – Neels
    Nov 22 at 20:13













up vote
0
down vote










up vote
0
down vote









The recurrence equation can actually be solved. The coefficient of $t^{2k+1}$ turns out to be $$ frac{3(-1)^k}{k!(2k-3)(4k^2-1)}$$
It turns out that the solution can be expressed in closed form:



$$ x(t) = frac{2t^3+5t}{8} e^{-t^2} + frac{sqrt{pi} text{erf}(t) (4 t^4 + 12 t^2 + 3)}{16}$$



although getting that from the series is not going to be easy.



EDIT:
Too bad you changed the initial conditions. Now the solution is as given by LutzL. The original was much more interesting, I think.






share|cite|improve this answer














The recurrence equation can actually be solved. The coefficient of $t^{2k+1}$ turns out to be $$ frac{3(-1)^k}{k!(2k-3)(4k^2-1)}$$
It turns out that the solution can be expressed in closed form:



$$ x(t) = frac{2t^3+5t}{8} e^{-t^2} + frac{sqrt{pi} text{erf}(t) (4 t^4 + 12 t^2 + 3)}{16}$$



although getting that from the series is not going to be easy.



EDIT:
Too bad you changed the initial conditions. Now the solution is as given by LutzL. The original was much more interesting, I think.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 23 at 20:26

























answered Nov 22 at 20:09









Robert Israel

317k23206457




317k23206457












  • Thanks for the reply. How do you figure this out so quickly?
    – Neels
    Nov 22 at 20:12










  • Also please check my edits, because i initially typed out the wrong limits.
    – Neels
    Nov 22 at 20:13


















  • Thanks for the reply. How do you figure this out so quickly?
    – Neels
    Nov 22 at 20:12










  • Also please check my edits, because i initially typed out the wrong limits.
    – Neels
    Nov 22 at 20:13
















Thanks for the reply. How do you figure this out so quickly?
– Neels
Nov 22 at 20:12




Thanks for the reply. How do you figure this out so quickly?
– Neels
Nov 22 at 20:12












Also please check my edits, because i initially typed out the wrong limits.
– Neels
Nov 22 at 20:13




Also please check my edits, because i initially typed out the wrong limits.
– Neels
Nov 22 at 20:13


















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