# Thread: Having trouble with a trig under. coeffecient

1. ## Having trouble with a trig under. coeffecient

im having trouble doing a trig coeffecient.

t'' + 2t' +6x = 10 cos 3t

i got the homogenius side, but now im having trouble with the right. Any help would be appreciated.

*edit*

I did get it to -3 A cos (3t) - 3Bsin (3t) - 6A sin (3t) + 6b cos (3t) = 10 cos 3t

2. Originally Posted by neven87
im having trouble doing a trig coeffecient.

t'' + 2t' +6x = 10 cos 3t

i got the homogenius side, but now im having trouble with the right. Any help would be appreciated.

*edit*

I did get it to -3 A cos (3t) - 3Bsin (3t) - 6A sin (3t) + 6b cos (3t) = 10 cos 3t
Something is really messed up in this equation. Are we solving for t(x) or x(t)? Is it possibly supposed to be
x'' + 2x' +6x = 10 cos 3t ?

-Dan

3. well actually equals d2x/dt2= t'' and dx/dt

dont know why i used t <shrug>

4. Originally Posted by neven87
well actually equals d2x/dt2= t'' and dx/dt

dont know why i used t <shrug>
Okay, so it's
$\displaystyle x^{\prime \prime} + 2x^{\prime} + 6x = 10 cos(3t)$

You said you have the homogeneous solution, so I'm not going to do that. I will simply say that the homogeneous solution contains no terms like Asin(3t) or Bcos(3t).

We need a particular solution. Given the form of the RHS I'm going to guess a particular solution
$\displaystyle x_p(t) = Asin(3t) + Bcos(3t)$

$\displaystyle x_p^{\prime}(t) = 3Acos(3t) - 3Bsin(3t)$

$\displaystyle x_p^{\prime \prime} = -9Asin(3t) - 9Bcos(3t)$

Inserting these into the differential equation gives:
$\displaystyle (-9Asin(3t) - 9Bcos(3t)) + 2(3Acos(3t) - 3Bsin(3t)) + 6(Asin(3t) + Bcos(3t))$ $\displaystyle = 10cos(3t)$

$\displaystyle (-9A - 6B + 6A)sin(3t) + (-9B + 6A + 6B)cos(3t) = 10cos(3t)$

The only way for this to be true for all t is for the coefficients of the trig functions to be the same on both sides of the equation. Thus:
$\displaystyle -9A - 6B + 6A = 0$
and
$\displaystyle -9B + 6A + 6B = 10$

The first equation says
$\displaystyle -3A - 6B = 0$ ==> $\displaystyle B = -\frac{1}{2}A$

Inserting this into the second equation gives:
$\displaystyle -9 \cdot -\frac{1}{2}A + 6A + 6 \cdot -\frac{1}{2}A = 10$ <-- Multiply both sides by 2

$\displaystyle 9A + 12A - 6A = 20$

$\displaystyle 15A = 20$

$\displaystyle A = \frac{20}{15} = \frac{4}{3}$

Thus
$\displaystyle B = -\frac{1}{2} \cdot \frac{4}{3} = -\frac{2}{3}$

$\displaystyle x_p(t) = \frac{4}{3}sin(3t) - \frac{2}{3}cos(3t)$

-Dan

5. Ah, made a mistake after the subing, thanks.