So, I was told solve the equation $y' - y = x^2$ using power series. Normal methods tell me that the solution is $y = c_{0}e^{x}-x^{2}-2x-2$, and this can be verified by plugging it back in. However, I am stuck on trying to solve this with power series.

We assume $y = \Sigma_{n=0}^{\infty}a_{n}x^{n}$. Thus, $y' = \Sigma_{n=0}^{\infty}a_{n+1}(n+1)x^{n}$. I plug these into the original equation.

$$\Sigma_{n=0}^{\infty}a_{n+1}(n+1)x^{n}-\Sigma_{n=0}^{\infty}a_{n}x^{n}=x^{2} \\ a_{1}x^{0}+2a_{2}x^{1}+3a_{3}x^{2}+4a_{4}x^{3}+\dots-a_{0}x^{0}-a_{1}x^{1}-a_{2}x^{2}-a_{3}x^{3}-\dots = x^{2}$$ By equating powers of $x$, I find the following relations $$a_{1}-a_{0}=0 \\ 2a_{2}-a_{1}=0 \\ 3a_{3}-a_{2}=1 \\ 4a_{4}-a_{3}=0 \\ \vdots \\ na_{n}-a_{n-1}=0 $$ So, I write the coefficients as $$ a_{1}=a_{0}\\ a_{2}=\frac{a_{1}}{2}=\frac{a_{0}}{2}\\ a_{3}=\frac{1}{3}+\frac{a_{2}}{3}=\frac{a_{0}}{6}+\frac{1}{3}\\ a_{4}=\frac{a_{0}}{24}+\frac{1}{12}\\ \vdots\\ a_{n}=\frac{a_{n}}{n!}+\frac{2}{n!} $$ Combining these to form $y$, I get $$y=\Sigma_{n=0}^{\infty}\frac{a_{0}}{n!}x^{n}+\Sigma_{n=3}^{\infty}\frac{2}{n!}x^{n} $$ The first bit gives me $a_{0}e^{x}$ as expected, but I don't see how to extract $-x-2x-2$ from the second half.

Is there something wrong in my approach that lead to an incorrect answer, or am I missing something in the manipulation of power series?

  • 3
    $\begingroup$ Your answer is correct. Notice $$\sum_{n=3}^\infty\frac2{n!}x^n=\sum_{n=0}^\infty\frac2{n!}x^n-\sum_{n=0}^2\frac2{n!}x^n=2e^x-x^2-2x-2$$ $\endgroup$ – Kemono Chen May 27 '18 at 5:00

Your approach is correct and you did no mistake

$$y=\sum_{n=0}^{\infty}\frac{a_{0}}{n!}x^{n}+\sum_{n=3}^{\infty}\frac{2}{n!}x^{n}$$ Transform the last part as an exponential and it will be aborbed by the first series excet for the three first terms $$\sum_{n=3}^{\infty}\frac{2}{n!}x^{n}=\sum_{n=0}^{\infty}\frac{2}{n!}x^{n} -2-2x-x^2$$ Therefore $$y=\sum_{n=0}^{\infty}\frac{a_{0}}{n!}x^{n}+\sum_{n=0}^{\infty}\frac{2}{n!}x^{n} -2-2x-x^2$$ $$y=\sum_{n=0}^{\infty}\frac{(a_{0}+2)}{n!}x^{n} -2-2x-x^2$$ $a_0$ is just a constant substitute $K=a_0+2$ $$y=\sum_{n=0}^{\infty}\frac{K}{n!}x^{n} -2-2x-x^2$$ $$y=K\sum_{n=0}^{\infty}\frac{x^{n}}{n!} -2-2x-x^2$$ Since $e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!} $ $$y={K}e^x -2-2x-x^2$$


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.