Solve Linear Inhomogeneous 2nd Order IVP: Joe Smalls' Question

[MarkFL]

Here is the question:

Solving this initial value problem?

So I'm trying to solve this:

y" + 4y = sin3t
y(0) =1
y'(0) = 17/5

I start by:

(D^2 + 4)(y) = sin3t
(D^2 + 9)(D^2 + 4)(y) = 0
y = C_1(cos3t) + C_2(sin3t) + C_3(cos2t) + C_4(sin2t)

Then I apply initial conditions:

1 = C_1 + C_3
17/5 = 3C_2 + 2C_4

After, I try to solve for a constant by plugging in the particular as y:

(D^2 + 9)(C_1(cos3t)) + C_2(sin3t)) = sin3t
-9C_1(cos3t) - 9C_2(sin3t) + 9C_1(cos3t) + 9C_2(sin3t) = sin3t
sin3t = 0

And that's where I'm stuck. The only thing I can think of is either that I made a mistake, or I can just assume C_1 = 0 because sin3t = 0 to make C_1(cos3t) + C_2(sin3t) = 0 be true. Any tips and help is appreciated!

If you must know, the answer is y = cos2t + 2sin2t - (1/5)(sin3t)

Also, if I can assume C_1 = 0, then I get C_3 = 1. But how do I solve for the other constants?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Re: Joe Smalls' question at Yahoo! Answers regarding a linear ingomogenous 2nd order IVP

Hello Joe Smalls,

We are given to solve:

\(\displaystyle y"+4y=\sin(3t)\) where \(\displaystyle y(0)=1,\,y'(0)=\frac{17}{5}\)

You have correctly applied the annihilator method and from this we may conclude that the homogenous solution is:

\(\displaystyle y_h(t)=c_1\cos(2t)+c_2\sin(2t)\)

And the particular solution is of the form:

\(\displaystyle y_p(t)=A\cos(3t)+B\sin(3t)\)

Using the method of undetermined coefficients, we begin by observing that:

\(\displaystyle y_p''(t)=-9y_p(t)\)

And so, by substitution into the ODE, we obtain:

\(\displaystyle -5A\cos(3t)-5B\sin(3t)=0\cdot\cos(3t)+1\cdot\sin(3t)\)

Comparison of coefficients gives us:

\(\displaystyle A=0,\,B=-\frac{1}{5}\)

And so our particular solution is:

\(\displaystyle y_p(t)=-\frac{1}{5}\sin(3t)\)

And thus the general solution by superposition is:

\(\displaystyle y(t)=y_h(t)+y_p(t)=c_1\cos(2t)+c_2\sin(2t)-\frac{1}{5}\sin(3t)\)

Hence:

\(\displaystyle y'(t)=-2c_1\sin(2t)+2c_2\cos(2t)-\frac{3}{5}\cos(3t)\)

Now, using the initial values, we obtain the system:

\(\displaystyle y(0)=c_1=1\)

\(\displaystyle y'(0)=2c_2-\frac{3}{5}=\frac{17}{5}\implies c_2=2\)

And so we find the solution satisfying the given IVP is:

\(\displaystyle y(t)=\cos(2t)+2\sin(2t)-\frac{1}{5}\sin(3t)\)