# Find the general solution

1. Apr 26, 2010

### Squeezebox

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

Find the general solution

y'' + 4y' +4y = 5xe^(-2x)

3. The attempt at a solution

I got (5/2)x^3*e^(-2x) as a particular solution. But I checked online at wolfram alpha and it says the particular solution is (5/6)x^3*e^(-2x). Using method of undetermined coefficients.

2. Apr 26, 2010

### rock.freak667

Solve the homogeneous part of y'' + 4y' +4y = 0. What are the roots of the characteristic equation?

3. Apr 26, 2010

### Squeezebox

The homogeneous part gave me

y=c1e-2x+c2xe-2x

from the root (D+2)2.

Multiplying by another (D+2) annihilated the 5xe-2x, resulting in a new solution

y=c1e-2x+c2xe-2x+c3x2e-2x

is that right?

4. Apr 26, 2010

### rock.freak667

Your right side is xe-2x, since -2 is seen twice in the homogeneous part, you must multiply it by x2.

Can you show your work on how you got the constant to be 5/2?

5. Apr 26, 2010

### Squeezebox

(D2+4D+4)(c3x2e-2x) = 5xe-2x

c3(4x2e-2x-8xe-2x+2e-2x+4(2xe-2x-2x2e-2x)+4x2e-2x) = 5xe-2x

Every thing but c3(2e-2x) reduces to zero

c3(2e-2x)=5xe-2x
c3=(5/2)x

6. Apr 26, 2010

### rock.freak667

I thought you got yp=c3x3e-2x

7. Apr 26, 2010

### Squeezebox

yp=c3x2e-2x
yp=(5/2)x*x2e-2x
yp=(5/2)x3e-2x