# solve the given differential by undetermined coefficients...

• Mar 5th 2011, 06:03 PM
slapmaxwell1
solve the given differential by undetermined coefficients...
ok so here is the problem

y'' +3y = -48(x^2)(e^3x)

ok so i get the first few couple of steps,

m^2 +3 = 0, m = +/- sqrt(3)i

so Yc = C1cos sqrt(3)x + C2sin sqrt(3)x

Yp = (Ax^2 +Bx + C)e^3x

so how do i find Yp? well let me ask first, is my setup right so far?

i need a lil help on this one....
• Mar 5th 2011, 06:06 PM
Ackbeet
Just plug Yp into the DE and equate like powers of x (the exponentials will cancel, I think). Where does that place you?
• Mar 5th 2011, 06:07 PM
Prove It
Quote:

Originally Posted by slapmaxwell1
ok so here is the problem

y'' +3y = -48(x^2)(e^3x)

ok so i get the first few couple of steps,

m^2 +3 = 0, m = +/- sqrt(3)i

so Yc = C1cos sqrt(3)x + C2sin sqrt(3)x

Yp = (Ax^2 +Bx + C)e^3x

so how do i find Yp? well let me ask first, is my setup right so far?

i need a lil help on this one....

Your setup looks fine so far.

Now since you know that \$\displaystyle \displaystyle y_p = (Ax^2 + Bx + C)e^{3x}\$, what are \$\displaystyle \displaystyle y'_p\$ and \$\displaystyle \displaystyle y''_p\$?
• Mar 5th 2011, 06:16 PM
slapmaxwell1
ok so Yp'= (2Ax + B)e^(3x) + 3(Ax^2+ Bx + C)e^(3x) and Yp'' = well its gonna be long :O)
• Mar 5th 2011, 06:20 PM
Prove It
Yes, but you need to evaluate it. Because the next step is to substitute \$\displaystyle \displaystyle y''_p\$ and \$\displaystyle \displaystyle y_p\$ into your DE, then simplify so that you can equate the coefficients of the like powers of \$\displaystyle \displaystyle x\$.
• Mar 5th 2011, 06:23 PM
slapmaxwell1
Yp''= 2Ae^(3x) + 3(2Ax + B)e^(3x) + 3(2Ax+B)e^(3x)+9(Ax^2+Bx+C)e^(3x) <----is this right?
• Mar 5th 2011, 06:27 PM
slapmaxwell1
ok i will evaluate it...so i substitute in the Y' and Y'' and then solve for A B and C
• Mar 5th 2011, 06:36 PM
Prove It
Is \$\displaystyle \displaystyle y'\$ part of your DE?