Thread: 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....

Just plug Yp into the DE and equate like powers of x (the exponentials will cancel, I think). Where does that place you?

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 , what are and ?

ok so Yp'= (2Ax + B)e^(3x) + 3(Ax^2+ Bx + C)e^(3x) and Yp'' = well its gonna be long :O)

Yes, but you need to evaluate it. Because the next step is to substitute and into your DE, then simplify so that you can equate the coefficients of the like powers of .

Yp''= 2Ae^(3x) + 3(2Ax + B)e^(3x) + 3(2Ax+B)e^(3x)+9(Ax^2+Bx+C)e^(3x) <----is this right?

ok i will evaluate it...so i substitute in the Y' and Y'' and then solve for A B and C

Is part of your DE?