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OpenStudy (anonymous):

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

OpenStudy (anonymous):

thats a y''

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

Give me a second to type this up :P

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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: \[y(t)=c_1 \cos(4t)+c_2 \sin(4t)+\frac{1}{32}t \sin(4t)\]

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

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

OpenStudy (anonymous):

did you do well in diff eq?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I'm also a math major though :P

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