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

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?
Reply 1
Avatar for vc94
vc94
12
-Acos(x)-Bsin(x) + 16(Acos(x)+Bsin(x))= 4cos(x)...do you mean?
Reply 2
Avatar for Zach827
Zach827
OP
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)
(edited 13 years ago)
Reply 3
Avatar for Josh_Dey
Josh_Dey
16
0.o
Reply 4
Avatar for Zach827
Zach827
OP
Original post by Josh_Dey

Original post by Josh_Dey
0.o


...?
Reply 5
Avatar for Sasukekun
Sasukekun
7
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.
Reply 6
Avatar for Zach827
Zach827
OP
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?
-Acos(x)-Bsin(x) = 4cos(x)

You didn't add 16y.
Reply 8
Avatar for Sasukekun
Sasukekun
7
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.
Reply 9
Avatar for Zach827
Zach827
OP
Oh! Okay, thanks. I'll try redoing it to see if I get the right answer.
Reply 10
Avatar for Nash24
Nash24
9
Sorry is this C4?
Reply 11
Avatar for Nobunoni
Nobunoni
Answer :-

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) (???) :smile:)) Tis been a very long time since' hehehe

+NNM+
(edited 13 years ago)
Reply 12
Avatar for ttoby
ttoby
15
Original post by Nash24

Original post by Nash24
Sorry is this C4?


No, you'd usually get this sort of question in 1st year uni maths.
Reply 13
Avatar for Nash24
Nash24
9
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.
Reply 14
Avatar for Zach827
Zach827
OP
It's for a Differential Equations class (book involves Linear Algebra).
Reply 15
Avatar for noobynoo
noobynoo
6
The solution is:

[br]y(x)=c1sin(4x)+c2cos(4x)+4cos(x)15[br][br]y(x) = c_1 sin(4 x)+c_2 cos(4 x)+\frac{4 cos(x)}{15}[br]

I just used a computer algebra to solve it. (i.e. I cheated) :colone:
Reply 16
Avatar for Zach827
Zach827
OP
Original post by noobynoo
The solution is:

[br]y(x)=c1sin(4x)+c2cos(4x)+4cos(x)15[br][br]y(x) = c_1 sin(4 x)+c_2 cos(4 x)+\frac{4 cos(x)}{15}[br]

I just used a computer algebra to solve it. (i.e. I cheated) :colone:


Lol, I know the solution >.< I was having trouble getting to it.... TI-89 calculators are amazing.

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