anonymous
  • anonymous
PLEASE HELP!!! Use power series to solve y''(t)=y*y'(t), y(0)=2 and y'(0)=-3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
You start with a series expansion for y\[y=\sum_{n=0}^{\infty}a_nt^n\] and take the necessary derivatives,\[y'=\sum_{n=0}^{\infty}na_nt^{n-1}\]and\[y''=\sum_{n=0}^{\infty}n(n-1)a_nt^{n-2}\]Since the first term of y' and first two terms of y'' equal zero, you have,\[y'=\sum_{n=1}^{\infty}na_nt^{n-1}\]and\[y''=\sum_{n=2}^{\infty}n(n-1)a_nt^{n-2}\]Now, because we want to add and multiply series and derive information on the coefficients, we need to ensure each sum expresses everything in terms of t^n. Hence, we transform n to n+1 in y' and n to n+2 in y''. If you do this, and expand, you'll see you haven't actually changed the sum, just the form. Doing this, and using your equation of y''=y*y', we have\[\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}t^n=\sum_{n=0}^{\infty}a_nt^n \times \sum_{n=0}^{\infty}(n+1)a_{n+1}t^n\]The last product is a Cauchy product (http://\[(n+1)(n+2)a_{n+2}=\sum_{k=0}^{n}(n-k+1)a_na_{n-k+1}\]en.wikipedia.org/wiki/Cauchy_product) which allows us to express the right-hand side as\[\sum_{n=0}^{\infty}\left( \sum_{k=0}^{n}(n-k+1)a_na_{n-k+1} \right)t^n\]Since the left-hand side and right-hand side are equal if and only if for each t^n the coefficients are equal,\[(n+1)(n+2)a_{n+2}=\sum_{k=0}^{n}(n-k+1)a_na_{n-k+1}\]That is,\[a_{n+2}=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}(n-k+1)a_na_{n-k+1}\]
Gina
  • Gina
o mr.lokisan i have a maths proces error here so i can not see the information
anonymous
  • anonymous
You use your initial conditions with the definition for y above to determine the first two coefficients you'll need in order to use that formula above to generate the rest.\[y(0)=\sum_{n=0}^{\infty}a_nt^n=a_0+a_1(0)+a_2(0)^2+...=a_0\]Since \[y(0)=2, a_0=2\]Similarly,\[y'(0)=\sum_{n=0}^{\infty}(n+1)a_{n+1}t^n=a_1+2a_2(0)+3a_3(0)^2+...=a_1\]So\[y'(0)=-3, a_1=-3\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Gina
  • Gina
o right but your information is writing maths error which means the symboys u used here ,,,so i can not see any formula plaese help by typing them here on my mail ;geo5phiri@yahoo.com
anonymous
  • anonymous
I thought this question was for omnideus. You need it too? I'll have to take screen shots and mail them. Some of your own symbols aren't coming through. Can I confirm your mail is geo5phiri@yahoo.com?
Gina
  • Gina
yes it is ,,,that is my mail
Gina
  • Gina
wow)) mr.lokisan am yo number fan,,could just teach me differentuation in general?? but i some basics
anonymous
  • anonymous
Hehe. Are my answers coming through for the other questions? Can you see the symbols?

Looking for something else?

Not the answer you are looking for? Search for more explanations.