Need help on Power Series Question

In summary, the student tried to solve for the coefficients of the homework equation but they didn't understand how to solve for them using initial conditions. They were able to solve for c0, c1, c2, ..., c5 by Differentiating the series expression.
  • #1
G4CKT
12
0

Homework Statement



y''+t^2*y'-y=1-t^2

Homework Equations



y(0)=-2
y'(0)=1

Find the first 6 coefficients

C1=-2
C2=1
C3=? (-1?)
C4=?
C5=?
C6=?

The Attempt at a Solution



Okay so I tried to do this but I'm not used to having anything on the RHS of the equation.

I got down to [tex]\Sigma[/tex](n+1)(n+2)an+2tn+[tex]\Sigma[/tex]an-1(n-1)tn-[tex]\Sigma[/tex]antn-a0+2*a2

Where the sigma's have n=2 on the bottoms of them. Don't know what to do next tho!
 
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  • #2
Set what you have equal to 1 - t^2.
Expand the summations on the left side up to about 5th degree terms.
Bring the two terms on the right side over to the left side.
Group all constants together, group all terms in t together, all terms in t^2 together, and so on up to the t^5 terms.
The expression on the left is identically equal to zero, so the constant term has to be zero, the coefficient of t has to be zero, and so on. These should allow you to solve for c0, c1, c2, ..., c5. (BTW, switch your coefficients from a_i to c_i so you don't confuse yourself.)
 
  • #3
Write the equation as

[tex] \sum [...] t^n = 1 - t^2 [/tex]

and then work with each n separately. For example, for n=0, you'd get something like 2a2 - a0 = 1, which gives you 2a2 with the help of the initial conditions. Then move to n=1, 2 ...

Alternatively, you can cheat a little; just looking at the differential equation you can immediately see what y''(0) has to be with those initial conditions. Then you can differentiate it once to find y'''(0) and so forth. Differentiating the series expression, you can immediately see the connection between the coefficients and derivatives at t=0.
 
  • #4
clamtrox said:
Write the equation as

[tex] \sum [...] t^n = 1 - t^2 [/tex]

and then work with each n separately. For example, for n=0, you'd get something like 2a2 - a0 = 1, which gives you 2a2 with the help of the initial conditions. Then move to n=1, 2 ...

Alternatively, you can cheat a little; just looking at the differential equation you can immediately see what y''(0) has to be with those initial conditions. Then you can differentiate it once to find y'''(0) and so forth. Differentiating the series expression, you can immediately see the connection between the coefficients and derivatives at t=0.

So for the n=0 you subbed the 0 into the t term as well?
 
  • #5
Bump!
 
  • #6
Nvm I got it guys thanks!
 

Related to Need help on Power Series Question

What is a power series?

A power series is an infinite series of the form Σan(x-c)n, where n is a non-negative integer, a is a constant coefficient, and c is a constant center. It is a type of mathematical series used to represent functions as a sum of powers of a variable.

What is the purpose of using a power series?

A power series is often used in mathematical analysis to approximate functions, especially when the function is difficult to evaluate directly. It can also be used to represent functions that cannot be expressed in terms of elementary functions.

How do I determine the convergence of a power series?

The convergence of a power series depends on the value of x and the coefficients an. There are several tests that can be used to determine convergence, such as the ratio test, root test, and integral test. The series will converge if the limit of these tests is less than 1.

What is the difference between a power series and a Taylor series?

A Taylor series is a specific type of power series that is centered at a specific point, usually the point where the function is being evaluated. It is used to approximate a function around that point, while a general power series can be used to approximate a function at any point.

How can I use a power series to find the value of a function?

To find the value of a function using a power series, you can substitute the value of x into the series and evaluate the resulting sum. However, this will only give an approximation of the function, as a power series is an infinite series and can only provide an estimate.

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