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Solve $$y''-3y'+2y=0$$ by using the power series method about the $x_0=0$.

My attempt:

The answer is $y(x)=Ae^x+Be^{2x}$

But how we can compute this by power series method? I suppose $y=\sum_{n=0}^{\infty}c_nx^n$

Therefore

$\sum_{n=2}^{\infty}n(n-1)c_nx^{n-2}-3\sum_{n=1}^{\infty}nc_nx^{n-1}+2\sum_{n=0}^{\infty}c_nx^n=0$

$\sum_{n=0}^{\infty}(n+2)(n+1)c_nx^n-3\sum_{n=0}^{\infty}nc_nx^n+2\sum_{n=0}^{\infty}c_nx^n=0$

I got the recurrence relation $$c_{n+2}=\frac{3(n+1)c_{n+1}-2c_n}{(n+1)(n+2)}$$ for $n=0,1,2,3,....$

But it gives

$c_2=\frac{3}{2}c_1-c_0,c_3=\frac{7}{6}c_1-c_0,c_4=\frac{5}{8}c_1-\frac{7}{12}c_0$ and so on.So the solution not matches with the answer.

Please help me thanks.

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  • $\begingroup$ still it is not clear that how I can get from here the answer $Ae^x+Be^{2x}$? $\endgroup$
    – Flip
    Dec 13, 2014 at 18:13
  • $\begingroup$ @JessicaK please check your calculations $(3/2)-(1/3)=7/6$ $\endgroup$
    – Flip
    Dec 13, 2014 at 18:34
  • $\begingroup$ Your value of $c_3$ is incorrect. $\endgroup$
    – Flip
    Dec 13, 2014 at 18:35
  • $\begingroup$ "But it gives $c_2=\frac{3}{2}c_1-c_0,c_3=\frac{7}{6}c_1-c_0,c_4=\frac{5}{8}c_1-\frac{7}{12}c_0$ and so on. So the solution not matches with the answer." Sorry but what do you mean by "the solution (does) not match the answer"? What does not match what? $\endgroup$
    – Did
    Dec 13, 2014 at 20:10

2 Answers 2

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To solve the recursion $$c_{n+2}=\frac{3(n+1)c_{n+1}-2c_n}{(n+1)(n+2)},$$ consider the change of variable $$a_n=n!\,c_n,$$ and note that $$a_{n+2}=3a_{n+1}-2a_n.$$ This is a linear recursion and the roots of the characteristic polynomial $x^2=3x-2$ are $x=1$ and $x=2$ hence $$a_n=\alpha+\beta 2^n,$$ where $(\alpha,\beta)$ solves the system $$\alpha+\beta=a_0=c_0,\qquad\alpha+2\beta=a_1=c_1.$$ Finally, for every $n$, $$c_n=\alpha\frac1{n!}+\beta\frac{2^n}{n!},$$ hence $$y(x)=\alpha\sum_{n=0}^\infty\frac{x^n}{n!}+\beta\sum_{n=0}^\infty\frac{2^nx^n}{n!}=\alpha\mathrm e^x+\beta\mathrm e^{2x}.$$

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  • $\begingroup$ The trick of change of variables is great! +1. $\endgroup$
    – voldemort
    Dec 14, 2014 at 21:31
  • $\begingroup$ @voldemort Thanks. Not that it is so difficult to guess... :-) $\endgroup$
    – Did
    Dec 15, 2014 at 10:30
  • $\begingroup$ ^ Ever so modest :). $\endgroup$
    – voldemort
    Dec 15, 2014 at 16:02
  • $\begingroup$ @voldemort Seriously? The recursion itself cries out for factorials and, if this would not suffice, one knows the final form of the solutions... $\endgroup$
    – Did
    Dec 15, 2014 at 16:04
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    $\begingroup$ @voldemort $\langle$ Blushes... $\rangle$ $\endgroup$
    – Did
    Dec 15, 2014 at 21:07
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Your work prior to the recurrence relation has some incorrect subscripts, and $c_{3} = c_{2} - (1/3)c_{1}$. You need to now rewrite your entire series in terms of $c_{0}$ and $c_{1}$ only, for example:

\begin{align*} c_{2} &= \frac{3}{2}c_{1} - c_{0}\\ c_{3} &= c_{2} - \frac{1}{3}c_{1}\\ c_{4} &= \frac{9}{12}c_{3} - \frac{2}{12}c_{2}. \end{align*}

This can be simplified into,

\begin{align*} c_{2} &= \frac{3}{2}c_{1} - c_{0}=\frac{3}{2!}c_1-\frac{2}{2!}c_0\\ c_{3} &= \left(\frac{3}{2}c_{1} - c_{0}\right) - \frac{1}{3}c_{1} = \frac{7}{6}c_{1}-c_{0} = \frac{7}{3!}c_1-\frac{6}{3!}c_0\\ c_{4} &= \frac{9}{12}\left(\frac{7}{6}c_{1}-c_{0}\right) - \frac{2}{12}\left( \frac{3}{2}c_{1} - c_{0}\right) = \frac{5}{8}c_1 - \frac{7}{12}c_0 = \frac{15}{4!}c_1 - \frac{14}{4!}c_0\\ \vdots \end{align*}

Now, recall that $y = \sum c_{n}t^{n}$, therefore

\begin{align}y &= c_{0} + c_{1}t + c_{2}t^{2} + \cdots \\ &= c_{0} + c_{1}t+ \left(\frac{3}{2!}c_1-\frac{2}{2!}c_0\right)t^{2} + \left(\frac{7}{3!}c_1-\frac{6}{3!}c_0\right)t^{3} + \left( \frac{15}{4!}c_1 - \frac{14}{4!}c_0\right)t^{4}+ \cdots \end{align}

Next, break up your series in the form

$c_{0}\sum_{n=0}^{\infty} a_{n}t^{n} + c_{1}\sum_{n=0}^{\infty} b_{n}t^{n}$

by pairing the powers of $t$'s together and regrouping the terms. This is pretty tedious.

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