Kaiya Hardin

2022-04-23

Power series solution of $y}^{\prime}-y={x}^{2$.

I try to get a power series solution.

$y\left(x\right)=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}{X}^{n},{y}^{\prime}\left(x\right)=\sum _{n=1}^{\mathrm{\infty}}n{a}_{n}{X}^{n-1}=\sum _{n=0}^{\mathrm{\infty}}(n+1){a}_{n+1}{X}^{n}$

$\sum _{n=0}^{\mathrm{\infty}}(n+1){a}_{n+1}{X}^{n}-\sum _{n=0}^{\mathrm{\infty}}{a}_{n}{X}^{n}={X}^{2}$

Then ${a}_{n+1}=\frac{{a}_{n}}{n+1}\mathrm{\forall}n\ne 2$

$n=2\Rightarrow 3{a}_{3}-{a}_{2}=1$

Then $a}_{1}={a}_{0},{a}_{2}=\frac{{a}_{0}}{2},{a}_{3}=\frac{2+{a}_{0}}{6},{a}_{4}=\frac{2+{a}_{0}}{4!},{a}_{5}=\frac{2+{a}_{0}}{24\cdot 4$

Then the solution is $y={a}_{0}+{a}_{0}{x}^{1}+\frac{{a}_{0}}{2}{x}^{2}+\frac{2+{a}_{0}}{6}{x}^{3}+\frac{2+{a}_{0}}{4!}{x}^{4}+\frac{2+{a}_{0}}{24\cdot 4}{x}^{5}\cdots$

Is it correct? Am I supposed to find $a}_{0$?

Jayla Matthews

Beginner2022-04-24Added 18 answers

Step 1

Your solution amounts to

$y-{a}_{0}(1+x+\frac{{x}^{2}}{2})=(2+{a}_{0})({\mathrm{e}}^{x}-1-x-\frac{{x}^{2}}{2})$.

Step 2

Notice you've indeed made a mistake. You have $a}_{n+1}=\frac{{a}_{n}}{n+1$, so $a}_{5}=\frac{{a}_{4}}{5}=\frac{2+{a}_{0}}{5!},{a}_{6}=\frac{{a}_{5}}{6}=\frac{2+{a}_{0}}{6!$, and so forth.

Back to the topic, $y=(2+{a}_{0}){\mathrm{e}}^{x}-2(1+x+\frac{{x}^{2}}{2})$.

It seems that any $a}_{0$ will fit $y}^{\prime}-y={x}^{2$. So just discard the foremost 2 and the solution is $y={a}_{0}{e}^{x}-2-2x-{x}^{2}$.

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