Particular Integral - Help equating terms

[_Greg_]

Homework Statement

Ok, so I have a 2nd order differential equation, I can get the complimentary function no problem, its getting numerical values for terms in the particular integral that I can't do.

Homework Equations

y'' - y' - 2y = t²

The Attempt at a Solution

Complimentiary function:

y(t) = Ae^2t + Be^-t

All fine and dandy, now particular interal for t²:

y_pi(t) = At² + Bt + C

Now we find the first and second order derivatives:

First: 2At + B

Second: 2A

Now substituting these terms back into original equation:

2A - (2At + B) - 2(At² + Bt + C) = t²

This is where I'm stuck, I'm looking at my notes for the next bit:

We can find A, B and C by equating terms, so:

t²: -2A -1
t: -2A - 2B = 0
1: 2A - b - 2C = 0

I don't understand that at all, can someone explain that a bit further?

 

[Pengwuino]

If you group up the terms, you have: (2A-B-2C)+(-2A-2B)t+(-2A)t^2 = t^2. You can also say you have (2A-B-2c)+(-2A-2B)t+(-2A)t^2 = 1t^2 + 0t + 0.

For that equation to be true, each set of coefficients must be equal, that is -2A = 1, -2A-2B = 0 aka 2A=-2B, and 2A-B-2C = 0. Three equations, three unknowns, you can solve for your particular solution.

 

[_Greg_]

Cheers Pengwuino, I think I get the general idea, so separate t² values from t's then everything else, so:

2A - (2At + B) - 2(At2 + Bt + C) = t²

Breaking down brackets:

2A - 2At - B -2At² - 2Bt - 2C = t²

(-2A)t² + (-2A -2B)t + (2A - B - 2C) = 1 + 0 + 0

Therefore:

(-2A) = 1
(-2A -2B) = 0
(2A - B - 2C) = 0

So,

A = -1/2
B = 1/2
C = 3/4

God this is sooo much to remember, find roots, use the correct general formula for the roots from memory, remember what form of particular integral to use, calculate 1st and 2nd derivatives, plug back into equation, break down and combine the complimentary function with the particular integral. Think i better get some practice!

E2A: I'v got a list of particular integral formats for different functions. This particular question I'm doing I though I better check, would I be right saying that for f(x) = x + 6 ------> y = Cx + D ?

 

[Pengwuino]

Do you mean you have a new problem where the inhomogeneous part is f(x) = x + 6? If so, yes, that is the particular solution in most cases. Remember, though, if your homogeneous solution has x or a constant as a solution, you need to tweak your attempt.