# Math Help - undetermined coeffecients...

1. ## undetermined coeffecients...

Im doing my DE and were going over complimentery and undetermined coeffeceints. Complimentery equations I could spout out all day long , but the undetermined coeffeceints is giving me some problems.

One of my main problems is im not quiet sure on what to subsitute for on the coeffeceints when the right side of the equations is a repeated root, or a trig function. Also i cant really ask the teacher as it is an online course. Some examples of my problems are:

PROBLEM 1)

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

i got the complimentery as yc= e^x(c1cos((squareroot(3))x)+C2sin(sqareroot 3))

but i go to build the coeffecients i get stuck. i have come up with A(e^4x)+ B(x(e^4x))+(c(x^2)(e^4x))

its suppose to come out :

y=(c1cos((squareroot(3))x)+C2sin(sqareroot 3))+((-4x^2)+4x-(4/3))(e^3x)

PROBLEM 2)

y(double der.) +4y = 3sin2x

im not sure what to expand this out to anymore...as when i set up the coeffecients i get (Acos(2x))+(bsin(2x)) which once i take the first and second derviate and subsitute i get all my coeffecients =0 which means i need to expand, but i am not sure what i would add to expand it....

I know this post is lengty and ugly but i would greatly appreciate any help or tips to get with the coeffecent building thanks!

2. #1:

$y''+3y=-48x^{2}e^{3x}$

From $m^{2}+3=0$ we find that $m_{1}=\sqrt{3}i$ and $m_{2}=-\sqrt{3}i$.

Then $y_{c}=C_{1}cos(\sqrt{3}x)+C_{2}sin(\sqrt{3}x)$

We can assume $y_{p}=(Ax^{2}+Bx+C)e^{3x}$

Sub into the DE we get:

$2A+6B+12C=0, \;\ 12A+12B=0, \;\ 12A=-48$

Solve the system and get:

Then, $A=-4, \;\ B=4, \;\ C=\frac{-4}{3}$

$y_{p}=(-4x^{2}+4x-\frac{4}{3})e^{3x}$

and therefore, hence, and whence:

$y=C_{1}cos(\sqrt{3}x)+C_{2}sin(\sqrt{3}x)+\left(-4x^{2}+4x-\frac{4}{3}\right)e^{3x}$

3. I might be asking a completly stupid question here:

The yp= (ax^2+bx+c)e^3x, the yp prime and yp double prime would be a long fun product rule?

4. Yes, for y'', you sub in a take the second derivative of $(Ax^{2}+Bx+C)e^{3x}$

which gives $(9Ax^{2}+12Ax+9Bx+2A+6B+9C)e^{3x}$

Then

$(9Ax^{2}+12Ax+9Bx+2A+6B+9C)e^{3x}+3(Ax^{2}+Bx+C)e^ {3x}$

= $(12Ax^{2}+12Ax+12Bx+2A+6B+12C)e^{3x}$

Now, equate coefficients and you get the system form my first post.

It's not that bad.

5. yeh, its not that bad...except i ussally make my errors on stupid mistakes i learned years ago