Question

Solve the differential equation y'+16y=1/4 cos4t

Answer 1

thats a y''

Answer 2

so y'' = 16y = 1/4 cos 4t,

D^2 + 16 = 1/4 cos 4t

y = C1 cos4t + c2 sin4t

(D^2+16)^2 = 0

If D^2+16 is squared, does the solution become like repeated roots, theres

C1 cos4t + c2 sin4t and then there's C3 x cos4t + C5 x sin4t?

Answer 3

Give me a second to type this up :P

Answer 4

Firstly, I would solve the homogeneous part:
y''+16y=0
So:
\[\lambda ^2+16=0 \rightarrow \lambda=\pm4i\]
Making the homogeneous part:
\[y_H(t)=c_1 \cos(4t)+c_2 \sin(4t)\]

Answer 5

Then you notice that you have cos(4t) in the non-homogeneous part so you have to include t to count for the redundancy.
So, using undetermined coefficients you would muster a guess of:
\[y_P(t)=At \cos(4t)+Bt \sin(4t)\]
Differentiating that twice you get:
\[y_P^''(t)=-8A \sin(4t)-16At \cos(4t)+8B \cos(4t)-16Bt \sin(4t)\]
If I did my differentiation right.
Plug that in and solve for A and B.
-8Asin(4t)-16Atcos(4t)+8Bcos(4t)-16Btsin(4t)+16Acos(4t)+16Btsin(4t)=(1/4)cos(4t)
Using linear algebra you get that:
A=0 and B=(1/32)
So your solution is:
\[y(t)=c_1 \cos(4t)+c_2 \sin(4t)+\frac{1}{32}t \sin(4t)\]

Answer 6

thank you

Answer 7

No problem, sorry it took me a minute to type up.

Answer 8

did you do well in diff eq?

Answer 9

I needed a 40 on the final exam to get an A. So I guess? :)

Answer 10

I'm also a math major though :P