# Differential Equation Help!

1. May 22, 2005

### Orion1

Find constants A,B and C such that the function:
$$y = Ax^2 + Bx + C$$

satisfies the differential equation:
$$y'' + y' - 2y = x^2$$

$$\frac{d}{dx} (y) = \frac{d}{dx} (Ax^2 + Bx + C) = 2Ax + B$$
$$y' = 2Ax + B$$

$$\frac{d}{dx} (y') = \frac{d}{dx} (2Ax + B) = 2A$$
$$y'' = 2A$$

$$2A + 2Ax + B - 2y = x^2$$

I have been assigned a problem that is not yet covered for another 7 chapters.

I do not understand the question...

Any suggestions?

2. May 22, 2005

### arildno

Why didn't you continue with the substitution:
$$2A+2Ax+B-2Ax^{2}-2Bx-2C=x^{2}$$

3. May 22, 2005

### Corneo

arildno is right. Here is a hint.

$$2A+2Ax+B-2Ax^{2}-2Bx-2C=x^{2}$$ can be cleaned up to

$$-2Ax^2+(2A-2B)x+2(A-C)+B = x^2$$

Do you see any terms $x$ terms on the RHS of the equation? What does this tells you about $2A-2B$? More over what should $-2A$ equal to, so it can satisfy the RHS of the equation? Apply the same idea for the terms $2(A-C)+B$

4. May 22, 2005

### dextercioby

You should regard what Arildno wrote as what it should really be,viz. an identity

$$(2A-2C+B)+(2A-2B)x-2Ax^{2}\equiv x^{2}$$

Daniel.

5. Jun 13, 2005

### murshid_islam

$$A = -\frac{1}{2}$$

$$B = -\frac{1}{2}$$

$$C = -\frac{3}{4}$$

Last edited: Jun 13, 2005