2nd Order nonhomogeneous ODE using Undetermined Coefficients

[Ramacher]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I know you have y_h 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 y_p1 I get y_p1=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=y_h+y_p1+y_p2 that y_p1 is going to be 0?

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

 

[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

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

 

[Ramacher]

roots of the characteristic polynomial are -3 and -2

The characteristic polynomial being the left side of the equation?

Isn't it a double root -3? (y+3)²?

 

[Mark44]

roots of the characteristic polynomial are -3 and -2,

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

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

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

 

[rock.freak667]

The characteristic polynomial being the left side of the equation?

Isn't it a double root -3? (y+3)²?

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

 

[Mark44]

The characteristic polynomial being the left side of the equation?

Isn't it a double root -3? (y+3)²?

Yes.

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

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

The general solution is y = y_h + y_p.