HELP!!! Calculus (differential equation) problem

[Scrapster]

Find the solution y= of:

y'' + y = tcos(2t)

Need to use "undetermined coefficients" on this one. That t in front of the cosine is screwing me up.

This is where I am:

(at + b)(dcos(2t) + esin(2t)) a-e are just constant coefficients with respect to t. I'm not sure what to do next. Do I distribute the (at + b), it gets pretty messy if I do.

anyone have suggestions?

 

[Passions]

Do dis:

r^2 + 1 =0 rite
=> r= +- i

Y(t)= (at + b)sin2t + (ct + d)cos2t

you know to do from there?

 

[DesignDawg]

DUH! The answer is 4. I thought it was pretty obvious. Dumbass. [Glad I never have to take another cal class]

Ricky
DesignDawg

 

[The Wildcard]

Man i took DiffEq last semester and i forgot already, lol.

 

[Scrapster]

Bobby, I'm already passed that. I found the yh, homogenous equation. Now i need to find, yp, the like form of: tcos(2t) (ex. (at + b)((cos2t) + sin2t)), then find the 2nd derivative of it to plug into my first original equation.

 

[Scrapster]

Help!

 

[Scrapster]

any suggestions?

 

[SnoopCat]

go to www.math.com

they have an online derivitive solver

 

[thEnEuRoMancER]

Homogeneous solution:

r^2+1=0
r=+-i

yh=d1*e^(it)+d2*e^(-ix)=c1*cos(t) + c2*sin(t)

Particular solution:

right side of equation: r(t)=t*cos2t is of the form r(t) = e^(at)*p(t)*cos(beta*t),
where a = 0, p(t)=t and beta=2

The particular solution for this form is:

yp=e^(ax)*sin(betax)*(Anxn+...+A1x+A0)+e^(ax)*cos(betax)*(Bnxn+...+B1x+B0)

the coefficients A1..An and B1...Bn are yet to be determined,

in our case a=0, beta=2, polinom p(t) is of the first degree, so:

yp=sin2t*(A1t+A0)+cos2t*(B1t+B0)

We must determine the coefficients A1,A0,B1 and B0 by deriving yp and inserting the derivatives into the differential equation:

yp" = sin2t*(-4A1t-4A0-4B1) + cos2t*(-4B1t-4B0+4A1)

so we get

y" + y = tcos(2t)

[sin2t*(-4A1t-4A0-4B1) + cos2t*(-4B1t-4B0+4A1)] + [sin2t*(A1t+A0)+cos2t*(B1t+B0)] = tcos(2t);

sin2t*(-3A1t-3A0-4B1) + cos2t*(-3B1t-3B0+4A1)] =tcos(2t)

the sin2t term on the left must equal 0, so

A1=0, -3A0-4B1=0, -3B1=1, -3B0+4A1=0

The coefficients are A1=0, B0=0, B1=-1/3, A0=4/9

The particular solution is therefore

yp=(4/9)sin2t-(1/3)tcos2t

And the general solution is

y=yh+yp= c1*cos(t) + c2*sin(t) + (4/9)sin2t-(1/3)tcos2t