Method of Undetermined Coefficients: y'' - 5y' + 6y = 3e^(3x) + 5x - 2

[Kevin98]

I need to use the method of undetermined coefficients to solve the equation:

y'' - 5y' + 6y = 3e^3x + 5x - 2

I'm having trouble with determining y_p for the right side of the equation.

Current Work:
am² + bm + c = 0
m² - 5m + 6 = 0
(m-2)(m-3) = 0
m = 2,3
y_h = C₁e^2x + C₂e^3x

y_p = Ae^3x + Bx + C
y'_p = 3Ae^3x + B
y''_p = 9Ae^3x

9Ae^3x - 5(3Ae^3x + B) + 6(Ae^3x + Bx + C) = 3e^3x + 5x - 2
9Ae^3x - 15Ae^3x - 5B + 6Ae^3x + 6Bx + 6C = 3e^3x + 5x - 2
9Ae^3x - 15Ae^3x + 6Ae^3x = 0
0e^3x = 0 ????
It doesn't seem like this step is correct.

 

[Deleted member 4993]

y'' - 5y' + 6y = 3e^3x + 5x - 2

I'm unsure of how to determine y_p for the right side of the equation.

Current work:
am² + bm + c = 0
m² - 5m + 6 = 0
(m-2)(m-3) = 0
m = 2,3
y_h = C₁e^2x + C₂e^3x

yp = Ae^3x + Bx + C
y'p = 3Ae^3x + B
y''p = 9Ae^3x

9Ae^3x - 5(3Ae^3x + B) + 6(Ae^3x + Bx + C) = 3e^3x + 5x - 2
9Ae^3x - 15Ae^3x - 5B + 6Ae^3x + 6Bx + 6C) = 3e^3x + 5x - 2
9Ae^3x - 15Ae^3x + 6Ae^3x = 0
0e^3x = 0???
It seems like representing 3e^3x as Ae³ is a mistake?

Since the homogeneous solution is also "forcing function" - the particular solution should be:

yp = A * x * e^3x + Bx + C

Now try it...

 

[Kevin98]

Thanks for the help, I got to the correct final answer!

How did you know to use the general form of Axe^3x? I was basing my decision on a video that explained the following:

Constant: A
Linear Polynomial: Ax + B
Quadratic: Ax² + Bx + C
Cubic: Ax³ + Bx² + Cx + D
sin (ax) or cos(ax): Acos(ax) + Bsin(ax)
e^ax: Ae^ax

Also, what would be the general form if the forcing function is 1 / (1+x²)? It doesn't seem to fall under any of the categories above.