Need help on Power Series Question

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Homework Help Overview

The problem involves solving a second-order differential equation with a power series approach. The equation is given as y'' + t^2*y' - y = 1 - t^2, with initial conditions y(0) = -2 and y'(0) = 1. The goal is to find the first six coefficients of the power series solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting the left-hand side equal to the right-hand side and expanding the summations to identify coefficients. There are suggestions to group terms by their powers of t and to apply initial conditions to derive relationships between coefficients. Some participants also mention differentiating the series to find connections between coefficients and derivatives at t=0.

Discussion Status

The discussion includes various approaches to manipulating the power series and applying initial conditions. Some participants offer guidance on how to organize the terms and derive coefficients, while others express uncertainty about specific steps in the process. There is a lack of explicit consensus on a single method, but multiple interpretations and strategies are being explored.

Contextual Notes

Participants note the challenge of having a non-zero right-hand side in the differential equation and the implications of the initial conditions on the coefficients. There is also mention of switching notation from a_i to c_i to avoid confusion.

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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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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.)
 
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.
 
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?
 
Bump!
 
Nvm I got it guys thanks!
 

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