0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ still it is not clear that how I can get from here the answer $Ae^x+Be^{2x}$? $\endgroup$ – Flip Dec 13 '14 at 18:13
  • $\begingroup$ @JessicaK please check your calculations $(3/2)-(1/3)=7/6$ $\endgroup$ – Flip Dec 13 '14 at 18:34
  • $\begingroup$ Your value of $c_3$ is incorrect. $\endgroup$ – Flip Dec 13 '14 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 '14 at 20:10
3
$\begingroup$

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 $$y(x)=\alpha\sum_{n=0}^\infty\frac{x^n}{n!}+\beta\sum_{n=0}^\infty\frac{2^nx^n}{n!}=\alpha\mathrm e^x+\beta\mathrm e^{2x}.$$

$\endgroup$
  • $\begingroup$ The trick of change of variables is great! +1. $\endgroup$ – voldemort Dec 14 '14 at 21:31
  • $\begingroup$ @voldemort Thanks. Not that it is so difficult to guess... :-) $\endgroup$ – Did Dec 15 '14 at 10:30
  • $\begingroup$ ^ Ever so modest :). $\endgroup$ – voldemort Dec 15 '14 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 '14 at 16:04
  • 1
    $\begingroup$ @voldemort $\langle$ Blushes... $\rangle$ $\endgroup$ – Did Dec 15 '14 at 21:07
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.