Question

Diff Eq.
Solve the initial value problem y'' + 9y = cos(3x) + sin(3x) for y(0) = 2, y'(0) = 1.

Answer 1

I just need the steps, not the actual solution. I found the imaginary solutions to be r=+-3i. What formula am I using, it should be different from the one you usually use for a homogeneous equation.

Answer 2

you have to find the "homogenous solution" and a particular solution...

Answer 3

\[r+\pm3 \]

Answer 4

this is slightly complicated, because the system is in resonance

Answer 5

Whats throwing me off is that the equation is equal to cos(3x) + sin(3x). Because is you look at attachment, I highlighted what would usually happen, if the equation was equal to something normal.

Answer 6

that method presents a slight complication in this particular problem
since the complementary solution and the terms on the right side are both sinusoids of the same angular velocity you have make some modifications...

Answer 7

you found the complementary solution?

Answer 8

I found a similar question, Example 4, at this link
http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-NonhomgenLinEqns_Stu.pdf

Answer 9

read "modification" thats mentioned right above example 6

Answer 10

thats what you have to apply in this case

Answer 11

Alright I'll try that, I have lecture right now...so I'll get back to you in about an hour and a half. Thanks for the help.

Answer 12

sure :)

Answer 13

go through example 5,
it's the same thing as the question you have posted,

if your trial solution and your complementary solution have similar terms, you have to multiply the trial solution by x.

Answer 14

in this case, your trial solution should be
y= Axcos(3x) + Bxsin(3x)