[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)=

Recall that

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

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

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 series should vanish and only the odd-indexed terms would remain.

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

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

• Even:

• Odd:

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 .

The even series vanishes, so that

for all integers k ≥ 1. But since , we find

and so on, which means the odd-indexed coefficients all vanish, .

This leaves us with the odd series,

We have

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

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

they are simplified

step-by-step explanation:

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.