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# Solve the Differential Equation y"+16y=4cos(x)? watch

1. I must find the general solution of the following differential equation:
y"+16y=4cos(x)

This is what I have:

Complimentary solution yc(x)
Solve for r:
r^2+16 = 0
r = ±4i
Plug into formula.
yc(x) = C1sin(4x)+C2cos(4x)

Particular test solution yt(x)
yt(x) = Acos(x)+Bsin(x)
y't(x) = Bcos(x)-A*sin(x)
y''t(x) = -Acos(x)-Bsin(x)
Set equal to RHS of original equation and solve for A and B:
-Acos(x)-Bsin(x) = 4cos(x)
A = -4, B = 0
Particular solution:
yp(x) = -4cos(x)

Final solution:
y(x) = yc(x) + yp(x)
y(x) = C1sin(4x)+C2cos(4x)-4cos(x)

I know the answer is supposed to be:
y(x) = C1sin(4x)+C2cos(4x)+(4/15)*cos(x)

So where did I go wrong with the particular solution? I think my particular test solution may be wrong, so how do I find a particular test solution?
2. -Acos(x)-Bsin(x) + 16(Acos(x)+Bsin(x))= 4cos(x)...do you mean?
3. I think that's a different path to solving it... I'm supposed to come up with a complimentary and particular solution and add those to get the total general solution.

In my notes there is an example where we solved:
y''+9y=2cos(3x)+3sin(3x)
The complimentary solution was
yc(x)=C1cos(3x)+C2sin(3x)
and the particular test solution was
yt(x)=x(Acos(3x)+Bsin(3x))
(Particular test solution multiplied by x due to similarity to complimentary solution.)
We then found the second derivative of that, set it equal to the original equation's RHS, and solved for A and B, then substituted those values in for the A and B in the original particular test solution to get the particular solution, which was added to the complimentary solution to get the final solution of:
y(x) = C1cos(3x)+C2sin(3x) - (1/2)xcos(3x)+(1/3)xsin(3x)
4. 0.o
5. (Original post by Josh_Dey)
0.o
...?
6. (Original post by Zach827)
Set equal to RHS of original equation and solve for A and B:
-Acos(x)-Bsin(x) = 4cos(x)
Not sure you did this bit right. I put it in and get A = 4/15 as your answer suggests.
7. (Original post by Sasukekun)
Not sure you did this bit right. I put it in and get A = 4/15 as your answer suggests.
Okay, but how?
-Acos(x)-Bsin(x) = 4cos(x)
B = 0
-Acos(x) = 4cos(x)
If A=-4
-(-4)cos(x) = 4cos(x)
4cos(x) = 4cos(x)

What did you do to get to 4/15?
8. -Acos(x)-Bsin(x) = 4cos(x)

9. (Original post by Zach827)
Okay, but how?
-Acos(x)-Bsin(x) = 4cos(x)
B = 0
-Acos(x) = 4cos(x)
If A=-4
-(-4)cos(x) = 4cos(x)
4cos(x) = 4cos(x)

What did you do to get to 4/15?
You seem to be missing the 16y.
10. Oh! Okay, thanks. I'll try redoing it to see if I get the right answer.
11. Sorry is this C4?

y = a cos(4x) + b sin(4x) + (4/15) cos(x)

y"+16y=4cos(x)

y'' + 16 y = 0

y_p = A cos(x) + B sin(x), and seek the coefficients A and B'

y = y_homogeneous + y_particular.

y = a cos(4x) + b sin(4x) + (4/15) cos(x) (???) )) Tis been a very long time since' hehehe

+NNM+
13. (Original post by Nash24)
Sorry is this C4?
No, you'd usually get this sort of question in 1st year uni maths.
14. (Original post by ttoby)
No, you'd usually get this sort of question in 1st year uni maths.
Oh thank god..
I have a C4 mock in a few days and most of this was not making any sense to me.
Thanks.
15. It's for a Differential Equations class (book involves Linear Algebra).
16. The solution is:

I just used a computer algebra to solve it. (i.e. I cheated)
17. (Original post by noobynoo)
The solution is:

I just used a computer algebra to solve it. (i.e. I cheated)
Lol, I know the solution >.< I was having trouble getting to it.... TI-89 calculators are amazing.

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Updated: March 28, 2011
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