Solving Non-homogeneous Systems of Differential Equations

[inquid]

Hello!

I am having trouble with a problem on my homework set. The general topic is on Non-Homogeneous Linear Systems of ODE. The subtopic is on Variation of Parameters for Non-Homogenous Linear Systems. Our instructor has attached the following text as a reference: https://math.libretexts.org/Bookshe..._Parameters_for_Nonhomogeneous_Linear_Systems.

---

The Problem

The problem reads:

Solve the system

$$x'+ \left[\begin{matrix} 6 & -2\\2&10 \end{matrix}\right]x=\left[\begin{matrix} e^{-t}\\0\end{matrix}\right]$$,

$$x_1(0)=1$$ and $$x_2(0)=-3$$
by the following steps.

1. Find $$e^{tP}$$.

$$e^{tP}=$$

2. Find $$\int^t_0 e^{sP} f(s) ds$$.

$$\int^t_0 e^{sP} f(s) ds=$$

3. Use the above to find the general solution. Recall that $$x_1(0)=1$$ and $$x_2(0)=-3$$.

$$x_1(t)=$$$$x_2(t)=$$

---

My Attempts

Now, I have tried solving for $$e^{tP}$$ by solving for the complementary system.

So, taking $$\left[\begin{matrix} 6 & -2\\2&10 \end{matrix}\right]$$, I insert the λ into the 1st and 4th positions.

Therefore, $$\left[\begin{matrix} 6-λ & -2\\2&10-λ \end{matrix}\right]$$.

Then, I solved for the eigenvalues by taking the determinant of the new matrix.

$$(6-λ)(10-λ)+4=64-16λ+λ^{2}=(λ-8)^{2}$$
Therefore, $$λ=8,8$$.

Now, to solve for the first eigenvector K, I take the new matrix and plug in 8 for λ.

Therefore, $$\left[\begin{matrix} -2 & -2\\2&2 \end{matrix}\right]$$.

Then, I took the first line of the matrix and set it equal to zero: $$-2k_1-2k_2=0$$.

Then, I set up the relation $$-2k_1=2k_2$$.

If I take $$k_1=1$$, then $$k_2=-1$$.

Therefore, my first eigenvector is $$\left[\begin{matrix} 1 \\ -1 \end{matrix}\right]$$.

Because this is a vector, I can scale the vector up by a factor of 2. Therefore, I have $$\left[\begin{matrix} 2 \\ -2 \end{matrix}\right]$$
To solve for my second eigenvector, I took the matrix $$\left[\begin{matrix} -2 & -2\\2&2 \end{matrix}\right]$$ and set it equal to the first eigenvector $$\left[\begin{matrix} 2 \\ -2 \end{matrix}\right]$$.

Therefore, I have the equation $$-2p_1-2p_2=2$$.

Then, I set up the relation $$-2p_1=2+2p_2$$. If $$p_2=1$$, then $$p_1=-2$$.

Therefore, the second eigenvector is $$\left[\begin{matrix} -2 \\ 1 \end{matrix}\right]$$.

Now that I have both eigenvectors, and knowing that this is a case of repeated eigenvalues, I can derive the general solution of the complementary system:

$$\left[\begin{matrix} x \\ y \end{matrix}\right]=c_1\left[\begin{matrix} 2 \\ -2 \end{matrix}\right]e^{8t} +c_2 \left( \left[\begin{matrix} 2 \\ -2 \end{matrix}\right]te^{8t}+\left[\begin{matrix} -2 \\ 1 \end{matrix}\right]e^{8t}\right)$$.

Now, this is where I am very, very lost.

The online program I am doing my homework on (MOER), can give me the answer key.

The answer they give for the first problem is:

$$e^{tP}=$$

$$(1-2t)e^{8t}$$	$$(-2t)e^{8t}$$
$$(2t)e^{8t}$$	$$(1+2t)e^{8t}$$

I do not know how to derive these answers. I first assumed I could take my solved general solutions above and just ignore the coefficients, but they do not yield what I want.

---

Plea for Help

I have been trying to work on these problems for almost a full week now, spending hours a day scouring the internet and textbooks, contacting my professor and going to math tutors but all to no avail.

Any help would be greatly appreciated.

Thank you in advance!

Inquid

 

[topsquark]

Hello!

I am having trouble with a problem on my homework set. The general topic is on Non-Homogeneous Linear Systems of ODE. The subtopic is on Variation of Parameters for Non-Homogenous Linear Systems. Our instructor has attached the following text as a reference: https://math.libretexts.org/Bookshelves/Differential_Equations/Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/10:_Linear_Systems_of_Differential_Equations/10.07:_Variation_of_Parameters_for_Nonhomogeneous_Linear_Systems.

---

The Problem

The problem reads:

Solve the system

$$x'+ \left[\begin{matrix} 6 & -2\\2&10 \end{matrix}\right]x=\left[\begin{matrix} e^{-t}\\0\end{matrix}\right]$$,

$$x_1(0)=1$$ and $$x_2(0)=-3$$
by the following steps.

1. Find $$e^{tP}$$.

$$e^{tP}=$$

2. Find $$\int^t_0 e^{sP} f(s) ds$$.

$$\int^t_0 e^{sP} f(s) ds=$$

3. Use the above to find the general solution. Recall that $$x_1(0)=1$$ and $$x_2(0)=-3$$.

$$x_1(t)=$$$$x_2(t)=$$

---

My Attempts

Now, I have tried solving for $$e^{tP}$$ by solving for the complementary system.

So, taking $$\left[\begin{matrix} 6 & -2\\2&10 \end{matrix}\right]$$, I insert the λ into the 1st and 4th positions.

Therefore, $$\left[\begin{matrix} 6-λ & -2\\2&10-λ \end{matrix}\right]$$.

Then, I solved for the eigenvalues by taking the determinant of the new matrix.

$$(6-λ)(10-λ)+4=64-16λ+λ^{2}=(λ-8)^{2}$$
Therefore, $$λ=8,8$$.

Now, to solve for the first eigenvector K, I take the new matrix and plug in 8 for λ.

Therefore, $$\left[\begin{matrix} -2 & -2\\2&2 \end{matrix}\right]$$.

Then, I took the first line of the matrix and set it equal to zero: $$-2k_1-2k_2=0$$.

Then, I set up the relation $$-2k_1=2k_2$$.

If I take $$k_1=1$$, then $$k_2=-1$$.

Therefore, my first eigenvector is $$\left[\begin{matrix} 1 \\ -1 \end{matrix}\right]$$.

Because this is a vector, I can scale the vector up by a factor of 2. Therefore, I have $$\left[\begin{matrix} 2 \\ -2 \end{matrix}\right]$$
To solve for my second eigenvector, I took the matrix $$\left[\begin{matrix} -2 & -2\\2&2 \end{matrix}\right]$$ and set it equal to the first eigenvector $$\left[\begin{matrix} 2 \\ -2 \end{matrix}\right]$$.

Therefore, I have the equation $$-2p_1-2p_2=2$$.

Then, I set up the relation $$-2p_1=2+2p_2$$. If $$p_2=1$$, then $$p_1=-2$$.

Therefore, the second eigenvector is $$\left[\begin{matrix} -2 \\ 1 \end{matrix}\right]$$.

Now that I have both eigenvectors, and knowing that this is a case of repeated eigenvalues, I can derive the general solution of the complementary system:

$$\left[\begin{matrix} x \\ y \end{matrix}\right]=c_1\left[\begin{matrix} 2 \\ -2 \end{matrix}\right]e^{8t} +c_2 \left( \left[\begin{matrix} 2 \\ -2 \end{matrix}\right]te^{8t}+\left[\begin{matrix} -2 \\ 1 \end{matrix}\right]e^{8t}\right)$$.

Now, this is where I am very, very lost.

The online program I am doing my homework on (MOER), can give me the answer key.

The answer they give for the first problem is:

$$e^{tP}=$$

$$(1-2t)e^{8t}$$	$$(-2t)e^{8t}$$
$$(2t)e^{8t}$$	$$(1+2t)e^{8t}$$

I do not know how to derive these answers. I first assumed I could take my solved general solutions above and just ignore the coefficients, but they do not yield what I want.

---

Plea for Help

I have been trying to work on these problems for almost a full week now, spending hours a day scouring the internet and textbooks, contacting my professor and going to math tutors but all to no avail.

Any help would be greatly appreciated.

Thank you in advance!

Inquid

You might find this to be illuminating.

-Dan

 

[mario99]

First, you sent the title for Variation of Parameters but then you solved with Matrix Exponential. Second, the integral is missing a lot of things, with traditional notation it should be like this:

$\displaystyle \bold{X} = e^{\bold{A}t}\bold{C} + e^{\bold{A}t}\int_{t_0}^{t}e^{\bold{-A}s}\bold{F}(s) \ ds$

 

[inquid]

Here is a screenshot of how the problem was written in MOER.

 

[fresh_42]

Where has your $e^{-t}$ term gone? The linear equation system is homogeneous in the second component. That allows us to eliminate $x_1(t)$ and get $0=32x_2+8x_2'+0.5x_2''+e^{-t} .$ By using your ansatz (almost) $x_2(t)=Ae^{-8t}+Bte^{-8t}+Ce^{-t}$ I got with $x_2(0)=-3=A+C$
$$x_2(t)=-\dfrac{149}{49}e^{-8t}+Bte^{-8t}+\dfrac{2}{49}e^{-t}$$Next, I would take this into the first component
$$x_1(t)= -5x_2(t)-0.5x_2'(t)$$and determine $B$ by $x_1(0)=1.$

 

[mario99]

Hello!
$$e^{tP}=$$

$$(1-2t)e^{8t}$$	$$(-2t)e^{8t}$$
$$(2t)e^{8t}$$	$$(1+2t)e^{8t}$$

I think that you are interested only in this part:

$\displaystyle \int_{t_0}^{t}e^{\bold{-A}s}\bold{F}(s) \ ds$

which should probably derive the table above. I did not check your $e^{\bold{-A}s}$ is correct or not. Therefore, we have to do all the work from the start.

 

[inquid]

Hello all:

Although I have not yet solved for the first two parts, I have managed to solve for the general equation (part three).

My first mistake was that I did not subtract the matrix with x from both sides.

My more grievous mistake was that I solved the IVP for the coefficients before I put in the particular solution. I used undetermined coefficients to solve for them.

Therefore, the correct general solutions are:

$$x_1(t)=\frac{145}{49}e^{-8t}-\frac{30}{7}te^{-8t}-\frac{30}{14}e^{-8t}+\frac{9}{49}e^{-t}$$
$$x_2(t)=\frac{-145}{49}e^{-8t}+\frac{30}{7}te^{-8t}-\frac{2}{49}e^{-t}$$.

 

[mario99]

$$e^{tP}=$$

$$(1-2t)e^{8t}$$	$$(-2t)e^{8t}$$
$$(2t)e^{8t}$$	$$(1+2t)e^{8t}$$

I do not know how to derive these answers.

My method gives me:

$\displaystyle e^{\bold{P}t} = e^{\bold{A}t} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\-2e^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}$

The reason my matrix is a little different than MOER because I used $\displaystyle \bold{A} = \begin{bmatrix}-6 & 2 \\ -2 & -10 \end{bmatrix}$

You did not move $\displaystyle \bold{A}$ to the right which is a standard way to solve the system of the first order differential equations and you used $\displaystyle \bold{A} = \begin{bmatrix}6 & -2 \\ 2 & 10 \end{bmatrix}$

If I do my calculations on $\displaystyle \bold{A} = \begin{bmatrix}6 & -2 \\ 2 & 10 \end{bmatrix}$, I get the same results as MOER.

Note: I have also solved the system, but I did not like to post my solution because it was not the purpose of the thread. There was only one little step left which was to use the initial condition. Was I lazy? Maybe!

-Mario, Dan's student

 

[mario99]

Correction. I forgot to multiply by $t$ in the first column, second row:

$\displaystyle e^{\bold{P}t} = e^{\bold{A}t} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}$

If you would like to learn the method of how to derive this table, tell me. But it depends on Inverse Laplace Transform as well as Inverse Matrix. If you are good at these, you will find it easy how to derive them.

 

[inquid]

Correction. I forgot to multiply by $t$ in the first column, second row:

$\displaystyle e^{\bold{P}t} = e^{\bold{A}t} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}$

If you would like to learn the method of how to derive this table, tell me. But it depends on Inverse Laplace Transform as well as Inverse Matrix. If you are good at these, you will find it easy how to derive them.

I would appreciate it if you showed me, please.

Thank you very much.

 

[topsquark]

-Mario, Dan's student

You are not my student.

-Dan

 

[mario99]

You are not my student.

-Dan

 

[mario99]

Where has your $e^{-t}$ term gone? The linear equation system is homogeneous in the second component. That allows us to eliminate $x_1(t)$ and get $0=32x_2+8x_2'+0.5x_2''+e^{-t} .$ By using your ansatz (almost) $x_2(t)=Ae^{-8t}+Bte^{-8t}+Ce^{-t}$ I got with $x_2(0)=-3=A+C$
$$x_2(t)=-\dfrac{149}{49}e^{-8t}+Bte^{-8t}+\dfrac{2}{49}e^{-t}$$Next, I would take this into the first component
$$x_1(t)= -5x_2(t)-0.5x_2'(t)$$and determine $B$ by $x_1(0)=1.$

Hi fresh_42,

My particular solution is $\displaystyle \begin{bmatrix}\frac{9}{49} \\[6pt] \frac{-2}{49} \end{bmatrix}e^{-t}$

While

Your particular solution is $\displaystyle \begin{bmatrix}1 \\[6pt] \frac{2}{49} \end{bmatrix}e^{-t}$

I am wondering who among us has made the mistake!

 

[mario99]

I would appreciate it if you showed me, please.

Thank you very much.

Which $\bold{A}$ do you want to use?

 

[fresh_42]

Hi fresh_42,

My particular solution is $\displaystyle \begin{bmatrix}\frac{9}{49} \\[6pt] \frac{-2}{49} \end{bmatrix}e^{-t}$

While

Your particular solution is $\displaystyle \begin{bmatrix}1 \\[6pt] \frac{2}{49} \end{bmatrix}e^{-t}$

I am wondering who among us has made the mistake!

My theoretical knowledge of LDE is more than a bit rusty so I have chosen a direct approach.

Here's my calculation for $x_2(t).$ I used $x=x_1$ and $y=x_2$ and $z',z''$ where I would normally write $\dot{z},\ddot{z}$ to make typing easier:

$$\begin{array}{lll} \begin{pmatrix}e^{-t}\\0\end{pmatrix}&= \begin{pmatrix}x'\\y'\end{pmatrix}+\begin{pmatrix}6&-2\\2&10\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}\\[10pt] 0&=y'+2x+10y \\ e^{-t}&=x'+6x-2y\\ x&=-5y-0.5y' \\ x'&=-5y'-0.5y''\\ e^{-t}&=-5y'-0.5y''-30y-3y'-2y=-32y-8y'-0.5y'' \end{array}$$
$$\begin{array}{lll} 0&=(6-t)(10-t)+4=t^2-16t+64=(t+8)^2\\ y&=Ae^{-8t}+Bte^{-8t}+Ce^{-t}\ ,\ A+C=-3\\ y'&=-8Ae^{-8t}+Be^{-8t}-8Bte^{-8t}-Ce^{-t}\\ y'&=(B-8A)e^{-8t}-8Bte^{-8t}-Ce^{-t}\\ y''&=-8(B-8A)e^{-8t}-8Be^{-8t}+64Bte^{-8t}+Ce^{-t}\\ y''&=(-16B+64A)e^{-8t}+64Bte^{-8t}+Ce^{-t} \end{array}$$
$$\begin{array}{lll} 0&=32y+8y'+0.5y''+e^{-t} \\ &=32Ae^{-8t}+32Bte^{-8t}+(32C+1)e^{-t}\\ &\phantom{=}+(8B-64A)e^{-8t}-64Bte^{-8t}-8Ce^{-t}\\ &\phantom{=}+(-8B+32A)e^{-8t}+32Bte^{-8t}+0.5Ce^{-t}\\ &=(24.5C-1)e^{-t}\text{ i.e. } C=\dfrac{2}{49} \text{ and }A=-\dfrac{149}{49} \end{array}$$
I haven't calculated $x_1(t),$ yet. Let's see:

$$\begin{array}{lll} x(t)&=-5y(t)-0.5y'(t)\\ &=-5Ae^{-8t}-5Bte^{-8t}-5Ce^{-t}+(-0.5B+4A)e^{-8t}+4Bte^{-8t}+0.5Ce^{-t}\\ &=(-A-0.5B)e^{-8t}-Bte^{-8t}-4.5Ce^{-t}\\ x(0)&=1=-A-0.5B-4.5C=\dfrac{140}{49}-0.5B \text{ and }B=\dfrac{26}{7} \end{array}$$
Hence my solution is

$$\begin{array}{lll} \begin{pmatrix}x\\y\end{pmatrix}&= \begin{pmatrix}A\\ -A- 0.5B \end{pmatrix}e^{-8t} +\begin{pmatrix}B\\ -B\end{pmatrix}te^{-8t} +\begin{pmatrix}C\\ -4.5C \end{pmatrix}e^{-t}\\ &=\dfrac{1}{49}\begin{pmatrix}-149\\ 58\end{pmatrix}e^{-8t} + \dfrac{1}{49}\begin{pmatrix}182\\ -182\end{pmatrix}te^{-8t} +\dfrac{1}{49}\begin{pmatrix}2\\ -9 \end{pmatrix}e^{-8t} \end{array}$$
Remains to check for sign errors, initial values, and derivatives. I estimate the possibility of a sign error by 5% and the initial values are correct. The check of whether my solution is a solution is easy but a bit troublesome.

 

[mario99]

Your method is not Matrix approach. Still it is unique. I will show my solution since you have shown yours.

Note: Errors may occur. We are humans.

When I do the OP method, I get:

$\displaystyle \bold{X} = c_1\begin{bmatrix}1 \\[5pt] -1 \end{bmatrix}e^{-8t} + c_2\left(\begin{bmatrix}1 \\[5pt] -1 \end{bmatrix} te^{-8t} + \begin{bmatrix}\frac{1}{2} \\[5pt] 0 \end{bmatrix}e^{-8t}\right) + \begin{bmatrix} \frac{9}{49} \\[5pt] -\frac{2}{49} \end{bmatrix}e^{-t}$

I was lazy to apply the initial condition, but I think I have to now. I get: $\displaystyle c_1 = \frac{145}{49}$ and $\displaystyle c_2 = -\frac{30}{7}$

$\displaystyle \bold{X} = \frac{145}{49}\begin{bmatrix}1 \\[5pt] -1 \end{bmatrix}e^{-8t} - \frac{30}{7}\left(\begin{bmatrix}1 \\[5pt] -1 \end{bmatrix} te^{-8t} + \begin{bmatrix}\frac{1}{2} \\[5pt] 0 \end{bmatrix}e^{-8t}\right) + \begin{bmatrix} \frac{9}{49} \\[5pt] -\frac{2}{49} \end{bmatrix}e^{-t}$


$\displaystyle \bold{X} = \frac{1}{49}\begin{bmatrix}40 \\[5pt] -145 \end{bmatrix}e^{-8t} + \frac{1}{7}\begin{bmatrix} -30 \\[5pt] 30 \end{bmatrix} te^{-8t} + \frac{1}{49}\begin{bmatrix} 9 \\[5pt] -2 \end{bmatrix}e^{-t}$

This guarantees that the initial condition is correct. Other things such as the derivatives have to be checked by CAS or AI.

I decided to use another method to solve the problem and this time I will use the Matrix Exponential.

$\displaystyle \bold{X} = e^{\bold{A}t}\bold{C} + e^{\bold{A}t}\int_{t_0}^{t}e^{\bold{-A}s}\bold{F}(s) \ ds$

I have already calculated $\displaystyle e^{\bold{A}t}$ in post #9.

$\displaystyle e^{\bold{A}t} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}$

And we know $\displaystyle \bold{F}(s) = \begin{bmatrix} 1 \\ 0 \end{bmatrix}e^{-s}$

This gives:

$\displaystyle \bold{X} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\[5pt]-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}\begin{bmatrix} C_1 \\[5pt] C_2 \end{bmatrix} + \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\[5pt]-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}\int_{t_0}^{t}\begin{bmatrix}(1 - 2s)e^{8s} & -2se^{8s} \\[5pt]2se^{8s} & (1 + 2s)e^{8s} \end{bmatrix}\begin{bmatrix} e^{-s} \\[5pt] 0 \end{bmatrix} \ ds$

Now we will focus on solving this integral:

$\displaystyle \int_{t_0}^{t}\begin{bmatrix}(1 - 2s)e^{8s} & -2se^{8s} \\[5pt]2se^{8s} & (1 + 2s)e^{8s} \end{bmatrix}\begin{bmatrix} e^{-s} \\[5pt] 0 \end{bmatrix} \ ds$


$\displaystyle = \int_{0}^{t}\begin{bmatrix}(1 - 2s)e^{7s} \\[5pt]2se^{7s} \end{bmatrix} \ ds$


$\displaystyle = \frac{1}{49}\begin{bmatrix}(9 - 14t)e^{7t} - 9 \\[5pt] (14t - 2)e^{7t} + 2 \end{bmatrix}$

Back to our solution:

$\displaystyle \bold{X} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\[5pt]-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}\begin{bmatrix} C_1 \\[5pt] C_2 \end{bmatrix} + \frac{1}{49}\begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\[5pt]-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}\begin{bmatrix}(9 - 14t)e^{7t} - 9 \\[5pt] (14t - 2)e^{7t} + 2 \end{bmatrix}$


$\displaystyle = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\[5pt]-2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}\begin{bmatrix} C_1 \\[5pt] C_2 \end{bmatrix} + \frac{1}{49}\begin{bmatrix} 9e^{-t} - 9e^{-8t} - 14te^{-8t} \\[5pt] -2e^{-t} + 2e^{-8t} + 14te^{-8t} \end{bmatrix}$


$\displaystyle = C_1\begin{bmatrix}(1 + 2t)e^{-8t} \\[5pt]-2te^{-8t} \end{bmatrix} + C_2\begin{bmatrix}2te^{-8t} \\[5pt] (1 - 2t)e^{-8t} \end{bmatrix} + \frac{1}{49}\begin{bmatrix} 9e^{-t} - 9e^{-8t} - 14te^{-8t} \\[5pt] -2e^{-t} + 2e^{-8t} + 14te^{-8t} \end{bmatrix}$

Continuation is in the next Post.

 

[mario99]

Applying the initial condition gives: $C_1 = 1 \$ and $\ C_2 = -3$

$\displaystyle = \begin{bmatrix}(1 + 2t)e^{-8t} \\[5pt]-2te^{-8t} \end{bmatrix} - 3\begin{bmatrix}2te^{-8t} \\[5pt] (1 - 2t)e^{-8t} \end{bmatrix} + \frac{1}{49}\begin{bmatrix} 9e^{-t} - 9e^{-8t} - 14te^{-8t} \\[5pt] -2e^{-t} + 2e^{-8t} + 14te^{-8t} \end{bmatrix}$

Collecting the same terms together gives:

$\displaystyle = \frac{1}{49}\begin{bmatrix}9 \\[5pt]-2 \end{bmatrix}e^{-t} + \frac{1}{49}\begin{bmatrix}49 - 9 \\[5pt] 2 - 147\end{bmatrix}e^{-8t} + \frac{1}{49}\begin{bmatrix}98 - 294 - 14 \\[5pt]-98 + 294 + 14 \end{bmatrix}te^{-8t}$


$\displaystyle = \frac{1}{49}\begin{bmatrix}9 \\[5pt]-2 \end{bmatrix}e^{-t} + \frac{1}{49}\begin{bmatrix}40 \\[5pt] -145\end{bmatrix}e^{-8t} + \frac{1}{49}\begin{bmatrix}-210 \\[5pt]210 \end{bmatrix}te^{-8t}$

Final solution is:

$\displaystyle \bold{X}= \frac{1}{49}\begin{bmatrix}9 \\[5pt]-2 \end{bmatrix}e^{-t} + \frac{1}{49}\begin{bmatrix}40 \\[5pt] -145\end{bmatrix}e^{-8t} + \frac{1}{49}\begin{bmatrix}-210 \\[5pt]210 \end{bmatrix}te^{-8t}$

Or

$\displaystyle x(t) = \frac{9}{49}e^{-t} + \frac{40}{49}e^{-8t} - \frac{210}{49}te^{-8t}$

$\displaystyle y(t) = -\frac{2}{49}e^{-t} - \frac{145}{49}e^{-8t} + \frac{210}{49}te^{-8t}$

 

[mario99]

My theoretical knowledge of LDE is more than a bit rusty so I have chosen a direct approach.

Here's my calculation for $x_2(t).$ I used $x=x_1$ and $y=x_2$ and $z',z''$ where I would normally write $\dot{z},\ddot{z}$ to make typing easier:

$$\begin{array}{lll} \begin{pmatrix}e^{-t}\\0\end{pmatrix}&= \begin{pmatrix}x'\\y'\end{pmatrix}+\begin{pmatrix}6&-2\\2&10\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}\\[10pt] 0&=y'+2x+10y \\ e^{-t}&=x'+6x-2y\\ x&=-5y-0.5y' \\ x'&=-5y'-0.5y''\\ e^{-t}&=-5y'-0.5y''-30y-3y'-2y=-32y-8y'-0.5y'' \end{array}$$
$$\begin{array}{lll} 0&=(6-t)(10-t)+4=t^2-16t+64=(t+8)^2\\ y&=Ae^{-8t}+Bte^{-8t}+Ce^{-t}\ ,\ A+C=-3\\ y'&=-8Ae^{-8t}+Be^{-8t}-8Bte^{-8t}-Ce^{-t}\\ y'&=(B-8A)e^{-8t}-8Bte^{-8t}-Ce^{-t}\\ y''&=-8(B-8A)e^{-8t}-8Be^{-8t}+64Bte^{-8t}+Ce^{-t}\\ y''&=(-16B+64A)e^{-8t}+64Bte^{-8t}+Ce^{-t} \end{array}$$
$$\begin{array}{lll} 0&=32y+8y'+0.5y''+e^{-t} \\ &=32Ae^{-8t}+32Bte^{-8t}+(32C+1)e^{-t}\\ &\phantom{=}+(8B-64A)e^{-8t}-64Bte^{-8t}-8Ce^{-t}\\ &\phantom{=}+(-8B+32A)e^{-8t}+32Bte^{-8t}+0.5Ce^{-t}\\ &=(24.5C-1)e^{-t}\text{ i.e. } C=\dfrac{2}{49} \text{ and }A=-\dfrac{149}{49} \end{array}$$
I haven't calculated $x_1(t),$ yet. Let's see:

$$\begin{array}{lll} x(t)&=-5y(t)-0.5y'(t)\\ &=-5Ae^{-8t}-5Bte^{-8t}-5Ce^{-t}+(-0.5B+4A)e^{-8t}+4Bte^{-8t}+0.5Ce^{-t}\\ &=(-A-0.5B)e^{-8t}-Bte^{-8t}-4.5Ce^{-t}\\ x(0)&=1=-A-0.5B-4.5C=\dfrac{140}{49}-0.5B \text{ and }B=\dfrac{26}{7} \end{array}$$
Hence my solution is

$$\begin{array}{lll} \begin{pmatrix}x\\y\end{pmatrix}&= \begin{pmatrix}A\\ -A- 0.5B \end{pmatrix}e^{-8t} +\begin{pmatrix}B\\ -B\end{pmatrix}te^{-8t} +\begin{pmatrix}C\\ -4.5C \end{pmatrix}e^{-t}\\ &=\dfrac{1}{49}\begin{pmatrix}-149\\ 58\end{pmatrix}e^{-8t} + \dfrac{1}{49}\begin{pmatrix}182\\ -182\end{pmatrix}te^{-8t} +\dfrac{1}{49}\begin{pmatrix}2\\ -9 \end{pmatrix}e^{-8t} \end{array}$$
Remains to check for sign errors, initial values, and derivatives. I estimate the possibility of a sign error by 5% and the initial values are correct. The check of whether my solution is a solution is easy but a bit troublesome.

FYI: The two different methods that I used have given the same solution which is also the same result that the OP has given in #7. Therefore, 99% my solution is correct!

 

[fresh_42]

FYI: The two different methods that I used have given the same solution which is also the same result that the OP has given in #7. Therefore, 99% my solution is correct!

Ok, I'll differentiate but before I have to correct that I confused $x,y$ in post #15 so I should have had (and let me get rid of that stupid denominator)

$$\begin{array}{lll} 49\begin{pmatrix}x\\y\end{pmatrix}&= \begin{pmatrix}58\\-149\end{pmatrix}e^{-8t} + \begin{pmatrix}-182\\ 182\end{pmatrix}te^{-8t} +\begin{pmatrix} -9 \\ 2\end{pmatrix}e^{-t} \end{array}$$and therefore
$$\begin{array}{lll} 49\begin{pmatrix}x'\\y'\end{pmatrix} &=\begin{pmatrix} -646\\ 1374\end{pmatrix}e^{-8t} +\begin{pmatrix}1456\\-1456\end{pmatrix}te^{-8t} +\begin{pmatrix} 9 \\ -2\end{pmatrix}e^{-t} \end{array}$$Now,
$$\begin{array}{lll} 49&\cdot\begin{pmatrix}6&-2\\2&10\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}\\ &=\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} 58 \\ -149 \end{pmatrix}e^{-8t} +\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} - 182 \\ 182 \end{pmatrix}te^{-8t} +\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} -9 \\ 2\end{pmatrix}e^{-t}\\ &=\begin{pmatrix} 646\\-1374 \end{pmatrix}e^{-8t} +\begin{pmatrix} -1456\\1456 \end{pmatrix}te^{-8t} +\begin{pmatrix}-58 \\ 2 \end{pmatrix}e^{-t} \end{array}$$and
$$\begin{array}{lll} 49&\cdot\begin{pmatrix}x'\\y'\end{pmatrix}+49\cdot\begin{pmatrix}6&-2\\2&10\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}\\ &=\begin{pmatrix}9\\-2\end{pmatrix}e^{-t}+\begin{pmatrix}-58 \\ 2 \end{pmatrix}e^{-t}=\begin{pmatrix}-49 \\ 0 \end{pmatrix}e^{-t} \end{array}$$D*****, a sign error in the $e^{-t}$ term or as Dirac had once pointed out: "There must be a sign error in an odd number of places."

A closer inspection revealed that I had $24.5C-1=0$ where I should have had $24.5C+1=0$ and thus $C=- \dfrac{2}{49}$ instead of the positive value in my calculation.

 

[mario99]

Ok, I'll differentiate but before I have to correct that I confused $x,y$ in post #15 so I should have had (and let me get rid of that stupid denominator)

D*****, a sign error in the $e^{-t}$ term or as Dirac had once pointed out: "There must be a sign error in an odd number of places."

A closer inspection revealed that I had $24.5C-1=0$ where I should have had $24.5C+1=0$ and thus $C=- \dfrac{2}{49}$ instead of the positive value in my calculation.

Thank you fresh_42 for making the effort to show where was your mistake. It's not really a mistake as you have already expected this in your post #15, sign errors may occur. At least now, we both agree with the same particular solution. The question remains how long are you going to take to agree that my complementary solution is correct?

The problem is that there is no quick way to check that!