# Differential Equation Help

1. May 20, 2007

1. The problem statement, all variables and given/known data
Find constants A, B, and C such that the function y = Ax^2+Bx+C satisfies the differential equation y''+y'-2y=x^2

2. The attempt at a solution

d/dx (y) = d/dx (Ax^2+Bx+C) = 2Ax+B
y' = 2Ax+B

d/dx (y') = d/dx (2Ax+B) = 2A
y'' = 2A

Now subbing back into the differential equation given:
(2A) + (2Ax+B) - 2y = x^2
2A + (2Ax+B) - 2(Ax^2+Bx+C) = x^2
2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

Moving the left hand side around:
(2A-2C+B) + (2A-2B)x - 2Ax^2 = x^2

However, now I have to solve for the constants and I'm not exactly sure how to figure that out. Any help would be great! Thanks guys :D

Last edited: May 20, 2007
2. May 20, 2007

### neutrino

Well, in the final equation, you have an x^2 term, an x term and a constant on the left, while you have only an x^2 term on the right. Now, based on this, what can you conclude about the coefficients on the left?

3. May 20, 2007

Well the coefficients on the left hand side if we were to rearrange it to
-2Ax^2 + (2A-2B)x + (2A-2C+B)
would be -2, 1, 1 which in a way is kind of similar to the equation:
-2x^2 + x+ 1 which is a polynomial. However, I'm not too sure if I should be moving the x^2 term from the right hand side over to the left hand side and solving for 0. Haha I'm sorry if I didn't really answer your question.

4. May 20, 2007

### neutrino

Okay, let me put in a different way...

Suppose ax^2+bx+1 = 3x^2+cx+d (a,b,c,d constants)

What can you say about a, b, c and d?

5. May 20, 2007

Hmmm I must not be seeing the logic behind this at all, and usually I would. Maybe I'm just having an off day or something. 12 years of schooling and 1 year of University is failing me today. The only thing I can think of would be that you are moving the right hand side over and subtracting it to be:
ax^2+bx+1 - 3x^2 - cx - d = 0
(a-3)x^2 + (b-c)x - d + 1 = 0

I feel really silly right now for not understanding because I know it is more likely really simple because the same principle for what you are asking will apply to my above question.

6. May 20, 2007

### D H

Staff Emeritus
The way to solve a problem like

$$a_nx^n + \cdots + a_1x + a_0 = A_nx^n + \cdots + A_1x + A_0$$

is to split this one equation into n equations by grouping like powers of x:

$$a_nx^n = A_nx^n$$
$$a_1x = A_1x$$
$$a_0 = a_0$$

from which $a_k = A_k$.

How does this work? Evaluate the expression at x=0. The constant terms are all that are left; they must be equal. The higher-order powers must be equal as well (prove recursively). You don't need the proof to solve the problem. Simply equate like terms.

Last edited: May 20, 2007
7. May 20, 2007

OHHH alright! Thank you so much both of you. That last explanation I think helped the most. I knew it could not be a hardship to figure it out but I made it seem harder than it should be. The keywords that helped me the most was "equate like terms" which I know is what neutrino was trying to state too.

Going back to my question, I'm finding that A = -1/2, B = -1/2 and C = -3/4 I believe. I'm hoping that is right anyway. :D

8. May 20, 2007

### D H

Staff Emeritus
Thanks.

Bingo!

9. May 20, 2007