# Thread: help with annihilator method

1. ## help with annihilator method

alright. Heres the problem:
1. $\displaystyle y''' + y'' = 8x^2$

Heres what I do:

$\displaystyle m^3 + m^2 = 0$

$\displaystyle m = 0, m = -1$

$\displaystyle D^3(D^3+D^2)y = D^38x^2$

$\displaystyle m^3(m^3 + m^2) = 0 auxillary equation$

$\displaystyle m^5(m+1) = 0$

$\displaystyle m = 0,0,0,0,0, -1$

$\displaystyle Yc = C1 + C2x + C3x^2 +C4+x^3 +C5x^4 + C6e^-^x$

The particular solution has the form:

$\displaystyle Yp = Ax^2 + Bx + C$

$\displaystyle Y'p = 2Ax + B$

$\displaystyle Y''p = 2A$

$\displaystyle Y'''p = 0$

Plugging in the particular solution into the DE gives:

$\displaystyle 2A = 8x^2$

The book says the answer is:

$\displaystyle y = C1 + C2x +C3e^-^x + \frac{2}{3}x^4 - \frac{8}{3}x^3 + 8x^2$

What im lost on(yes its algebra) on how they get 8/3, 2/3, and 8. And how they knew where to put them.

Anyone willing to help me out? Thanks!!!

2. The homogeneous equation is

$\displaystyle y'''+y''=0 \iff D^3y+D^2y=0 \iff D^2(D+1)y=0$

So the solution to the homogeneous solution is

$\displaystyle y_c=c_1+c_2x+c_3e^{-x}$

Now to find the particular solution we

$\displaystyle D^2(D+1)y=8x^2 \iff D^5(D+1)y=0$

So the solution to this new equation are

$\displaystyle A+Bx+Cx^2+Dx^3+Ex^4+Fe^{-x}$

Now we need to removed any terms that are in the complimentary solution this gives

$\displaystyle y_p=Cx^2+Dx^3+Ex^4$

Now just plug this in to the ODE and this gives

$\displaystyle (6D+24Ex)+(2C+6Dx+12Ex^2)=8x^2$

$\displaystyle 12Ex^2+(24E+6D)x+(6D+2C)=8x^2$

This gives the system of equation

$\displaystyle 12E=8 \quad 24E+6D=0 \quad 6D+2C=0$

Then just plug these back into the particular solution.

3. Thank you. You are the man!