Question

[infinity]
Substitute y(x)= Σ 2 anx^n and the Maclaurin series for 6 sin3x into y' - 2xy = 6 sin 3x and equate the coefficients of like powers of x on both sides of the equation to n= 0. Find the first four nonzero terms in a power series expansion about x = 0 of a general
n=0
solution to the differential equation.

У(Ñ)= ___________

Answer 1

Recall that

$\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}$

Differentiating the power series series for y(x) gives the series for y'(x) :

$y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n$

Now, replace everything in the DE with the corresponding power series:

$y'-2xy = 6\sin(3x) \implies$

$\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}$

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed $(n=2k)$ series should vanish and only the odd-indexed $(n=2k+1)$ terms would remain.

Split up both series on the left into even- and odd-indexed series:

$y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}$

$-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)$

Next, we want to condense the even and odd series:

• Even:

$\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}$

$=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}$

$=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}$

$=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}$

$=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}$

$=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}$

• Odd:

$\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}$

$=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}$

$=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}$

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that $a_1=0$.

The even series vanishes, so that

$(2k+1)a_{2k+1} - 2a_{2k-1} = 0$

for all integers k ≥ 1. But since $a_1=0$, we find

$k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0$

$k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0$

and so on, which means the odd-indexed coefficients all vanish, $a_{2k+1}=0$.

This leaves us with the odd series,

$\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$

$\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}$

We have

$k=0 \implies 2a_2 - 2a_0 = 6$

$k=1 \implies 4a_4-2a_2 = -1$

$k=2 \implies 6a_6-2a_4 = \dfrac1{20}$

$k=3 \implies 8a_8-2a_6 = -\dfrac1{840}$

So long as you're given an initial condition $y(0)\neq0$ (which corresponds to $a_0$), you will have a non-zero series solution. Let $a=a_0$ with $a_0\neq0$. Then

$2a_2-2a_0=6 \implies a_2 = a+3$

$4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4$

$6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}$

and so the first four terms of series solution to the DE would be

$\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}$