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$y'''+4y''+5y'+2y=10cos(t)$

Where it is subject to a condition

$y''(0)=3, y'(0)=0, y(0)=0$

Ok I'm having a very hard time to solving this b/c if one is going to solve using system of linear different equations don't we need a forcing term. I have solved for the Yc which is

$Y_c=c_1v_1e^{-t}+c_2(v_1+v_2t)e^{-t}+c_3v_3e^{-2t} $

and my lambda= -1,-1,-2

and my vectors are $v_1=[1,-1,1] v_2=[1,0,-1] v_3=[1,-2,4]$

but for the Yp I don't know how solve without forcing terms.

Am I doing something wrong?

Help would be much appreciated.

Thank You!

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  • $\begingroup$ Is that $y(0) = 0$? $\endgroup$ – Amzoti Jun 30 '13 at 1:23
  • $\begingroup$ yes sorry! y(0)=0 $\endgroup$ – bob Jun 30 '13 at 1:25
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Jun 30 '13 at 13:13
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I will guide you through hints and see if you can fill in the blanks using two methods and recommend a third.

Method 1 (Complimentary and Particular Solution)

For the complementary solution, we have:

$m^3 + 4m^2 + 5 m = 2 = 0 \rightarrow m_1 = -1, m_2 = -1, m_3 = -2$

This gives us a complimentary solution (when using the ICs) of:

$$y_c = 3 e^{-t} (t-1) + 3 e^{-2 t}$$

To find the particular solution, we guess at a solution in the form of:

$$y_p = a \cos t + b \sin t$$

Differentiate and substitute into ODE and solve for constants.

You should end up with:

$$y(t) = e^{-2 t} \left(-2 e^t (t-1)+2 e^{2 t} \sin t-1\right)-\cos t $$

Method 2 (Matrix DEQ)

We need to rewrite the system as a set of first order equations, so we let $x_1=y$ and have:

  • $x_1' = y' = x_2$
  • $x_2' = y'' = x_3$
  • $x_3' = y''' = -4y'' - 5y' -2y + 10 \cos t = -4x_3 - 5x_2 -2x_1 + 10 \cos t$

Our initial conditions are now expressed as $x_1(0) = 0, x_2(0) = 0, x_3(0) = 3$. Please make a note that we have time $t_0 = 0$ as we need that for later.

Now, we have a system of three equations and we can rewrite it using matrix equations as:

$$x'(t) = Ax(t) + F(t), \text{with IC vector}~~ C$$

To solve this matrix equation with forcing function we have:

$$\tag 1 \displaystyle x(t) = e^{At}C + \int_{t_0}^{t} e^{A(t-s)}F(s)~ds$$

So, we have:

$$\tag 2 x'(t) = \begin{bmatrix}x_1'(t)\\x_2'(t)\\x_3'(t) \end{bmatrix} = \begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\ -2 & -5 & -4\end{bmatrix}\begin{bmatrix}x_1(t)\\x_2(t)\\x_3(t) \end{bmatrix} + \begin{bmatrix}0\\0\\ 10\cos t \end{bmatrix}, C = \begin{bmatrix}0\\0\\3 \end{bmatrix}, t_0 = 0$$

We have a generalized eigenvector for $v_3$ due to eigenvalue with multiplicity of two. We have eigenvalues/eigenvectors as (I changed the order you used, but it does not matter):

  • $\lambda_1 = -2, ~v_1 = (1, -2, 4)$
  • $\lambda_2 = -1, ~v_2 = (1, -1, 1)$
  • $\lambda_3 = -1, ~v_3 = (2, -1, 0)$

Let me kick start you, to solve $(2)$, from $(1)$, we get:

$$e^{At}C = \begin{bmatrix}3 e^{-t} (t-1)+3 e^{-2 t}\\-3 e^{-t} (t-2)-6 e^{-2 t}\\3 e^{-t} (t-3)+12 e^{-2 t} \end{bmatrix}$$

Next do the integral (it is not bad), so see if you can do the others from this since you know the result.

Method 3: Variation of Parameters w/Greene's Function

This method is very useful because it makes it very easy to change the forcing function with little work.

Give it a go if you are bored!

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  • $\begingroup$ I'm liking your approach (+). $\endgroup$ – Namaste Jun 30 '13 at 1:36
  • $\begingroup$ Thanks so much for your input @Amzoti $\endgroup$ – bob Jun 30 '13 at 1:42
  • $\begingroup$ I do understand how to solve this using that method, but I need to solve this problem using the matrix approach. Sorry for the bad grammar! I'm embarrassed. $\endgroup$ – bob Jun 30 '13 at 1:57
  • $\begingroup$ Wow, I didn't even know you could solve for the forcing term, but I still don't understand how you did that. Is that some kind of formula that I should remember? $\endgroup$ – bob Jun 30 '13 at 2:17
  • $\begingroup$ @bob: Yes, it is very useful, but there are a lot of moving parts to this method, so Method 1 is preferred, maybe followed by Variation of Parameters, followed by this method and others! You can see these approaches in books on Control Theory for example. $\endgroup$ – Amzoti Jun 30 '13 at 2:26

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