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I'm given the following:

$$\begin{cases}\frac{dx}{dt} = t^2x\\ x(0) = 1\end{cases}$$

I'm asked to determine the taylor expansion for the solution to the $t^{10}$ term.

$$x(t) = a_0 + a_1 t + a_2 t^2 + a_3 t^3 + a_4 t^4 + a_5 t^7 + a_8 t^8 + a_9 t^9 + a_{10} t^{10} + o(t^{10})$$

How do I go about doing this? Any help at all would be greatly appreciated. The full problem is below as well.

The problem

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  • $\begingroup$ Hint: The answer is $$\color{red}{\bf 0}.$$ $\endgroup$ – Did Apr 13 '15 at 13:03
  • $\begingroup$ Maybe this isn't what they want, but the easiest thing would be just to solve the equation. $\endgroup$ – GPerez Apr 13 '15 at 13:28
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Assuming $x(t)=\sum_{i=0}^{n}a_i t^i$

Put this equation to the differential equation.

$$\frac{dx}{dt}=t^2x \Longrightarrow \sum_{i=0}^{n}i\ a_i\ t^{i-1}=t^2\sum_{i=0}^{n}a_i t^i $$

$$\Longrightarrow \sum_{i=0}^{n}i\ a_i\ t^{i-1}=\sum_{i=0}^{n}a_i t^{i+2} \Longrightarrow (i+3)a_{i+3}=a_i$$

For finding initial condition ($a_0, a_1, a_2$), we have to take a look at $x(0)=1$ and $\sum_{i=0}^{n}i\ a_i\ t^{i-1}=\sum_{i=0}^{n}a_i t^{i+2}$

$$\Longrightarrow x(0)=a_0=1$$ $$ 0\cdot a_0 t^{-1}+a_1+2a_2t+3a_3t^2\cdots =a_0t^2+a_1t^3+a_2t^4+\cdots \Longrightarrow a_1=a_2=0 $$

Q.E.D. We could obtain $a_{3k}=\frac{1}{3^k k!}$ by solving a recursion equation $(i+3)a_{i+3}=a_i$ with initial conditions $a_0=1,a_1=0,a_2=0$.

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You are asked to find out $x(t=0)$, $\frac{\mathrm{d}x}{\mathrm{d}t}(t=0)$, $\frac{\mathrm{d}^2x}{\mathrm{d}t^2}(t=0)$, and so on until the $10-th$ order. You know the value $x(t=0)=1$, and for the other terms in the expansion, you need to know $\frac{\mathrm{d}^kx}{\mathrm{d}t^k}(t=0)$ for $k\leq10$ ; but $\frac{\mathrm{d}^kx}{\mathrm{d}t^k}(t)=\frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}}\frac{\mathrm{d}x}{\mathrm{d}t}(t)=\frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}}(t^2x(t))$, so all you need to do is to differentiate this last expression (take care that $t^2$ AND $x(t)$ must be differentiated!). After it, take $t=0$ ; hence the term $a_k$ is $\frac{1}{k!}\frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}}(t^2x(t))$ at $t=0$ for $1\leq k \leq 10$.

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    $\begingroup$ You mean, differentiate $k$ times? Not sure I would have the courage... $\endgroup$ – Did Apr 13 '15 at 12:30
  • $\begingroup$ Yeah fixed it! You are fast :-) $\endgroup$ – Nicolas Apr 13 '15 at 12:31
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If $${d x\over dt}=t^2x$$ then $${d^2 x\over dt^2}={d \over dt}\left({d x\over dt}\right)=2tx+t^2{d x\over dt}=2tx+t^4x$$ and $${d^3 x\over dt^3}={d \over dt}\left({d^2 x\over dt^2}\right)=2x+2t{d x\over dt}+4t^3x+t^4{d x\over dt}=2x+2t^3x+4t^3x+t^6x = 2x+6t^3x+t^6x$$.

You can carry on a similar process for the others, but you'll need to put the value $t=0$ and the value for $x(t=0)$ into all these, and you should get a set of numbers which can be put into the Taylor expansion formula (remember to include the factorials on the bottom!)

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  • $\begingroup$ why the downvote?! $\endgroup$ – danimal Apr 13 '15 at 14:59

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