# 2nd Order nonhomogeneous ODE using Undetermined Coefficients

1. Aug 4, 2009

### Ramacher

1. The problem statement, all variables and given/known data

Find General Solution:
y"+6y'+9y=e-3x-27x2

2. Relevant equations

3. The attempt at a solution

I know you have yh which is the general solution to the left side of the equation set to 0 and then fine the particular solution.

When i try to find yp1 I get yp1=Ae-3x, y'p1=-3Ae-3x, and y"p1=9Ae-3x

Substituting that into the left side we get
9Ae-3x+6(-3Ae-3x)+9(Ae-3x)=Ae-3x
Canceling out Ae-3x we get
9A-18A+9A=1
0A=1

SO does this mean when i'm finding the general solution y=yh+yp1+yp2 that yp1 is gonna be 0?

I Know this is a very simple question but my mind is running in circles!

2. Aug 4, 2009

### rock.freak667

roots of the characteristic polynomial are -3 and -2, so your PI for the e-3x is not Ae-3x but Axe-3x, the PI for -27x2 is Bx2+Cx+D, so combining those two, your PI would be

$$y_{PI} = Axe^{-3x}(Bx^2+Cx+D)$$

3. Aug 4, 2009

### Ramacher

The characteristic polynomial being the left side of the equation?

Isn't it a double root -3? (y+3)2?

4. Aug 4, 2009

### Staff: Mentor

No, -2 is not a root of the characteristic equation. The only root is -3, and this is a double root.

5. Aug 4, 2009

### rock.freak667

oh right it is, I did that wrong in my head...in that case, Axe-3x would be Ax2e-3x

6. Aug 4, 2009

### Staff: Mentor

Yes.

Your solution to the homogeneous equation is yh = Ae-3x + Bxe-3x.

The particular solution to the nonhomogeneous problem is yp = Cx2e-3x + D + Ex + Fx2.

The general solution is y = yh + yp.