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The ODE below is required to help compute the coefficients of function. There isnt any information about this topic in my textbook so i am just wondering how i would go about this question?

In this question you will construct the series solution of the following ODE $dy/dx = 2y$

Of course we are making a big assumption here – that the solution $y(x)$ can be written as a Taylor series.

(a) Assume that $y(0) = A$ for some given number $A$. Use the ODE to compute the successive derivatives of $y(x)$ at $x = 0$.

(b) Use the result of part (a) to deduce the coefficients $a_0, a_1, a_2, a_3$ and an in the infinite series $y(x)=a_0 +a_1x+a_2x^2 +a_3x^3 +···+a_nx^n$

(c) Use the result of part (b) to write down $y(x)$ as an infinite series. Your answer should contain only one constant $A$ and should include the general term in the infinite series.

(d) Solve the ODE using any other method. Does this solution agree with what you found in part (c)? Justify your answer.

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you can differentiate the equation repeatedly to get the following: $$y' = 2y, y'' = 2y' = 2^2y, y^{(3)}= 2^3y, \cdots, y^{n} = 2^ny, \cdots \tag 1$$

suppose $y(0) = a.$ then substituting $x = 0,$ in $(1),$ we get $$y'(0) = 2a, y''(0)=2^2 a,\cdots,y^{(n)} = 2^n a, \cdots \tag2 $$

now using the maclaurin series $$y =y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+ \cdots+ \frac{x^n}{n!}y^{(n)}(0)+\cdots$$ which gives us $$\begin{align}y &=a+2ax+2^2\frac{x^2}{2!}a+ \cdots+ 2^n\frac{x^n}{n!}a+\cdots \\&=a\left( 1+2x+\frac{(2x)^2}{2!}+ \cdots+ \frac{(2x)^n}{n!}+\cdots \right)\\ &= ae^{2x}\end{align}$$

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Assume $y = \sum_{i=0}^{\infty}a_nx^n$ is a solution. Then $y' = \frac{dy}{dx} = \sum_{i=1}^{\infty} na_nx^{n-1}$ Now plug this into the ODE and try and combine the two series by manipulating the index on the first term. Once you do this you should be able to factor out an $x^n$ to get a recurrence relation to help you solve the rest of the problem! Try and find a pattern in this relation.

Try this, post your results and well help you further along.

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Series solution of a first order differential equation? Also you have written $dy/dy = 2y$:

Shurely shome mishtake mish Moneypenny?

EDIT Why downvote? I know it's fine to use series solution for first order; wlog that is; but it's not really needed. Also the typo is important.

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  • $\begingroup$ series solutions are perfectly fine for first order. also - I think its clear they meant $\frac{dy}{dx}$ $\endgroup$ – faith_in_facts May 14 '15 at 11:49

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