Problem with differential equation

[Sonal7]

I am unable to solve part c)
I have the right equation for x from part a and b, and rather than show you the workings heres the answer. I am not substituting in 0 for derivate of x as the T.P has to be when dx/dt =0.
So now I get
[MATH]0=-1/6e[SUP]-3t[/SUP] -4te[SUP]-3t[/SUP] -1/18 cos 3t[/MATH]Its a 2 mark Q so I am not sure where it needs a lot of work.

 

[MarkFL]

We are given the ODE:

[MATH]\frac{d^2x}{dt^2}+6\frac{dx}{dt}+9x=\cos(3t)[/MATH]
Now, concerning the homogeneous solution, we observe that the characteristic equation has the repeated root:

[MATH]r=-3[/MATH]
And so the homogeneous solution must therefore be:

[MATH]x_h(t)=c_1e^{-3t}+c_2te^{-3t}[/MATH]
Now for the particular solution, we cans use the method of undetermined coefficients and state:

[MATH]x_p(t)=A\cos(3t)+B\sin(3t)[/MATH]
Thus:

[MATH]x_p'(t)=-3A\sin(3t)+3B\cos(3t)[/MATH]
[MATH]x_p''(t)=-9A\cos(3t)-9B\sin(3t)[/MATH]
Substitution into the ODE gives us eventually:

[MATH]18B\cos(3t)-18A\sin(3t)=1\cos(3t)+0\sin(3t)[/MATH]
Equating like coefficients, we find:

[MATH](A,B)=\left(0,\frac{1}{18}\right)[/MATH]
Hence:

[MATH]x_p(t)=\frac{1}{18}\sin(3t)[/MATH]
And so by the principle of superposition, we find the general solution to the ODE is:

[MATH]x(t)=x_h(t)+x_p(t)=c_1e^{-3t}+c_2te^{-3t}+\frac{1}{18}\sin(3t)[/MATH]
Now, in order to use the initial values to determine the parameters, we compute:

[MATH]x'(t)=-3c_1e^{-3t}+c_2(e^{-3t}-3te^{-3t})+\frac{1}{6}\cos(3t)[/MATH]
And so using:

[MATH]x(0)=\frac{1}{2},\,x'(0)=0[/MATH]
We obtain the system:

[MATH]x(0)=c_1=\frac{1}{2}[/MATH]
[MATH]x'(0)=-3c_1+c_2+\frac{1}{6}=0\implies c_2=\frac{4}{3}[/MATH]
And so the solution to the given IVP is:

[MATH]x(t)=\frac{1}{2}e^{-3t}+\frac{4}{3}te^{-3t}+\frac{1}{18}\sin(3t)=\frac{8t+3}{6e^{3t}}+\frac{1}{18}\sin(3t)[/MATH]
This is equivalent to your result. So far, so good. Now, to answer part (c), Let's consider (found from the derivative we computed aboe and using the parameter values we found):

[MATH]x'(t)=-\frac{3}{2e^{3t}}+\frac{4(1-3t)}{3e^{3t}}+\frac{1}{6}\cos(3t)=0[/MATH]
Let's multiply by \(6e^{3t}\) to get:

[MATH]-9+8(1-3t)+e^{3t}\cos(3t)=0[/MATH]
Arrange as:

[MATH]e^{3t}\cos(3t)-24t-1=0[/MATH]
We aren't going to be able to solve for \(t\) explcitly, so we'll have to use a root finding technique like the Newton-Raphson method, or technology. Are you familiar with the Newton-Raphson method?

 

[Sonal7]

Its a 2 mark question. I am relieved that I got to the last line same as you and when there is lots of working I guess that I am doing too much. I think there is a - in the exponent e^-3t. I got the drift someone else that if t is large that makes that term 0. I am unsure about the rest. NR is on the agenda. But my brain will fry if i tried that but why not? Please lets try the shortcut and i will try NR later Oh no i am wrong, i can see whats going on.

 

[Sonal7]

[MATH]0=1/6 cos 3t[/MATH] but that doesnt work. dunno

 

[MarkFL]

Let's look at Newton's method, which is recursively defined as:

[MATH]t_{n+1}=t_{n}-\frac{f(t_{n})}{f'(t_{n})}[/MATH]
We have:

[MATH]f(t)=e^{3t}\cos(3t)-24t-1[/MATH]
Hence:

[MATH]f'(t)=3e^{3t}(\cos(3t)-\sin(3t))-24[/MATH]
And so we have the recursion:

[MATH]t_{n+1}=t_{n}-\frac{e^{3t_n}\cos(3t_n)-24t_n-1}{3e^{3t_n}(\cos(3t_n)-\sin(3t_n))-24}[/MATH]
Now, let's let \(t_0=31\) and we then find through successive computations:

[MATH]t_1\approx30.916403049103989863393159693[/MATH]
[MATH]t_2\approx30.8939130027647479035062935806260[/MATH]
[MATH]t_3\approx30.89233525175866335483577126330557608[/MATH]
[MATH]t_4\approx30.892327760467994341654295445992287175415[/MATH]
[MATH]t_5\approx30.8923277602996335116343627095320805794798717[/MATH]
[MATH]t_6\approx30.89232776029963351154932660224844502797447100742[/MATH]
[MATH]t_7\approx30.892327760299633511549326602248445027952777588797055[/MATH]
At this point I would be comfortable with the level of accuracy, and state:

[MATH]t\approx30.892327760299633511549326602248445028[/MATH]
For comparison, Desmos gives this value as:

[MATH]t=\frac{59\pi}{6}\approx30.89232776029963351154932660224844502793883242718854057292...[/MATH]
And so I would give the approximate location of \(A\) as:

[MATH](30.892327760299633511549326602248445028,-0.0555555555555555555555555555555555555)[/MATH]

 

[Sonal7]

I think this my revision for NR done. That is the right method but I wondered how they did it.