Question

Use power series to solve y''(t)=y*y'(t), y(0)=2 and y'(0)=-3

I know to assume that y=a0+a1t+a2t^2+a3t^3+a4t^4 (where the number following a is its subscript and not a coefficient on t)

and I know to get the first and second derivatives from there. What I am having trouble with is foiling the series for y and the series for y'. Could someone please help me finish this problem? If I could just multiply the two series together without any mistakes, I'd pretty much be done.

Answer 1

I answered your question earlier.

Answer 2

thank you so much for that

Answer 3

So it's okay?

Answer 4

Would you be willing to help me foil out the two series? My teacher wants me to show them out to the fourth degree (something I forgot to write into my original question)

Answer 5

the series for y and y' that is

Answer 6

Do you mean your teacher wants you to expand to the fourth degree?

Answer 7

yeah. I actually just did the foil for the fourth degree of y and y' . y*y'=a0a1+(2a0a2+(a1)^2)t+(3a0a3+3a1a2)t^2+(4a0a4+4a1a3+2(a2^2)t^3+(5a0a5+5a1a4+5a2a3)t^4. Here's my new problem though: Now that I have all of the values for them, I am told to equate different coefficients of powers of t, but I keep on screwing up the scratchwork. would you help me euate the different powers of t?

Answer 8

\[y \prime=10^{x+y}\]

Answer 9

I'm not sure how your reply makes sense here Gina.

Answer 10

That looks right. I haven't completed a total expansion, though. So, you're now needing to actually determine the coefficients? Are you allowed to use the formula I derived earlier?

Answer 11

\[a_{n+2}=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}(n-k+1)a_na_{n-k+1}\]

Answer 12

Yeah, but I think he just wants us to equate the different coefficients of different powers of t and solve algebraically. So for instance, the constant term for y''=yy' is 2a2=a0a1-->a2=(1/2)a0a1. then you go down the list and just put everything in terms of a0 and a1.

Answer 13

Gina, post your question again separately and I'll help.

Answer 14

does that make sense?

Answer 15

Yep. You have a0 and a1...the rest is just turgid algebra. I get the feeling you're screwing it up because it's tedious and has many 'moving parts'. It sounds like you know what to do, though. I don't know what else I can do.

Answer 16

Yeah, it sounds like your teacher just wants you to go through the motions.

Answer 17

help me with the actual algebra part? I really need to get this question done and all that I'm screwing up is the tedious part.

Answer 18

I'll see if I can get you started.

Answer 19

thank you so much. I think I can get the first few t terms without trouble. It's like the t^3 and t^4 terms taht are giving me trouble.

Answer 20

any luck?

Answer 21

I expanded first to check, which took some time.

Answer 22

\[2a_2=a_0a_1\rightarrow a_2=\frac{2 \times -3}{2}=-3\]
\[6a_3=2a_0a_2+a^2_1\rightarrow a_3=\frac{2 \times 2 \times -3+(-3)^2}{6}=-\frac{1}{2}\]

Is this looking familiar?

Answer 23

\[12a_4=3a_0a_3+3a_1a_2\]\[\rightarrow a_4=\frac{3*2*-\frac{1}{2}+3*-3*-3*-3}{12}=-7\]

Answer 24

\[20a_5=4a_0a_4+4a_1a_3+2a^2_2\rightarrow a_5=-\frac{8}{5}\]

Answer 25

\[30a_6=5a_0a_5+5a_1a_4+5a_2a_3 \rightarrow a_6=\frac{193}{60}\]