Solving Non-homogeneous Systems of Differential Equations

[fresh_42]

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!

Your first OP-like solution is not really different from my explicit calculation. It's just written a bit differently, more compact, and without my stupid sign error.

Your second solution with the integral is more interesting. I couldn't find how you derived the matrix $e^{\mathbf{A}t}$ which is the crucial part of all. I'm particularly interested in the procedure of exponentiation. I doubt that you have used BCH. Is there a nice way to do it?

And if you ended up with the correct solution, why do you still want to check the calculations?

 

[mario99]

The main purpose of this thread is to show the OP how to get $e^{\bold{A}t}$ which he has written it as $e^{\bold{P}t}$. The OP has not replied to this thread since yesterday and the only thing that I can think of is that he is having difficulties of accessing this math website as it happened to me before. (Or he got what he wanted from somewhere else.)

For the sake of completeness, I will show the method in this post. To find $e^{\bold{A}t}$, you need two matrices, $\bold{A}$ our main matrix and $\bold{I}$ the identity matrix. You also need a little skill in finding inverse matrices as well as finding inverse Laplace Transform. You don't need to be super in inverse Laplace Transform as everything you need is available in the inverse Laplace Transform table.

$\displaystyle \bold{A} = \begin{bmatrix}-6 & 2 \\-2 & -10 \end{bmatrix} \$ and $\displaystyle \ \bold{I} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \$

The first step is to solve this matrix:

$s\bold{I} - \bold{A}$

where $s$ is the parameter of the Laplace Transform.

$s\bold{I} - \bold{A} = \displaystyle s\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix}-6 & 2 \\-2 & -10 \end{bmatrix} = \displaystyle \begin{bmatrix}s & 0 \\ 0 & s \end{bmatrix} + \begin{bmatrix}6 & -2 \\2 & 10 \end{bmatrix} = \begin{bmatrix} s+ 6 & -2 \\2 & s + 10 \end{bmatrix}$

The next step is to find the inverse of the matrix above:

$\displaystyle (s\bold{I} - \bold{A})^{-1} = \begin{bmatrix}\frac{s + 10}{s^2 + 16s + 64} & \frac{2}{s^2 + 16s + 64} \\[7pt] \frac{-2}{s^2 + 16s + 64} & \frac{s + 6}{s^2 + 16s + 64} \end{bmatrix} = \begin{bmatrix}\frac{s + 10}{(s + 8)^2} & \frac{2}{(s + 8)^2} \\[7pt] \frac{-2}{(s + 8)^2} & \frac{s + 6}{(s + 8)^2} \end{bmatrix}$

Now you need to take inverse Laplace Transform of each element of the matrix above. With a little or no manipulation of each element, you can get the inverse Laplace Transform from the table.

And finally, we get what the OP was asking for:

$e^{\bold{P}t} = e^{\bold{A}t} = \begin{bmatrix}(1 + 2t)e^{-8t} & 2te^{-8t} \\[7pt] -2te^{-8t} & (1 - 2t)e^{-8t} \end{bmatrix}$

 

[mario99]

Your second solution with the integral is more interesting. I couldn't find how you derived the matrix $e^{\mathbf{A}t}$ which is the crucial part of all. Is there a nice way to do it?

Well there is also another method to find $e^{\bold{A}t}$ but it's not perfect with all matrices.

$\displaystyle e^{\bold{A}t} = \sum_{n=0}^{\infty}\bold{A}^n \frac{t^n}{n!}$

It works with any $\bold{A}$ that is $n \times n$ matrix. But it works perfectly in $n \times n$ diagonal matrices. It can work in the OP matrix but because it is not a diagonal matrix, we will have a hard time until we recognize the pattern produced by the infinite series. Therefore, it's better to use the inverse Laplace Transform to avoid the headache of thinking.

I doubt that you have used BCH.

What is BCH?

And if you ended up with the correct solution, why do you still want to check the calculations?

I had a doubt 1% that my solution was not correct, but thanks for confirming the 100% success.

 

[fresh_42]

Well there is also another method to find $e^{\bold{A}t}$ but it's not perfect with all matrices.

$\displaystyle e^{\bold{A}t} = \sum_{n=0}^{\infty}\bold{A}^n \frac{t^n}{n!}$

It works with any $\bold{A}$ that is $n \times n$ matrix. But it works perfectly in $n \times n$ diagonal matrices. It can work in the OP matrix but because it is not a diagonal matrix, we will have a hard time until we recognize the pattern produced by the infinite series. Therefore, it's better to use the inverse Laplace Transform to avoid the headache of thinking.

Guess, I should review the Laplace transformation calculus.

What is BCH?

Baker-Campbell-Hausdorff formula to correct the difference $e^{X}e^{Y} - e^{X+Y}.$ It is the formal reason for what you called a "headache of thinking".

I had a doubt 1% that my solution was not correct, but thanks for confirming the 100% success

My reviewed solution (and final confirmation of the correct result) is, for the sake of completeness:

1.) Elimination of $x=x_1$ and $x'=x'_1$ :

$$\begin{array}{lll} \begin{pmatrix}e^{-t}\\0\end{pmatrix}&= \begin{pmatrix}x'\\y'\end{pmatrix}+\begin{pmatrix}6&-2\\2&10\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}\, , \,x(0)=1\, , \,y(0)=-3\\[16pt] 0&=y'+2x+10y \\ e^{-t}&=x'+6x-2y\\ x&=-5y-0.5y'\\ x'&=-5y'-0.5y''\\ e^{-t}&=-5y'-0.5y''-30y-3y'-2y=-32y-8y'-0.5y'' \end{array}$$
2.) Ansatz to solve for $y=x_2$ :

$$\begin{array}{lll} 0&=(6-t)(10-t)+4=t^2-16t+64=(t+8)^2\\[16pt] y&=Ae^{-8t}+Bte^{-8t}+Ce^{-t}\ ,\ A+C=-3\\ y'&=-8Ae^{-8t}+Be^{-8t}-8Bte^{-8t}-Ce^{-t}\\ y'&=(B-8A)e^{-8t}-8Bte^{-8t}-Ce^{-t}\\ y''&=-8(B-8A)e^{-8t}-8Be^{-8t}+64Bte^{-8t}+Ce^{-t}\\ y''&=(-16B+64A)e^{-8t}+64Bte^{-8t}+Ce^{-t} \end{array}$$
3.) Calculation of the coefficients $A,C$ from the initial value $y(0)=-3$ :

$$\begin{array}{lll} 0&=32y+8y'+0.5y''+e^{-t}\\ &=32Ae^{-8t}+32Bte^{-8t}+(32C+1)e^{-t}\\ &\phantom{=}+(8B-64A)e^{-8t}-64Bte^{-8t}-8Ce^{-t}\\ &\phantom{=}+(-8B+32A)e^{-8t}+32Bte^{-8t}+0.5Ce^{-t}\\ &=(24.5C+1)e^{-t}\text{ i.e. } C=-\dfrac{2}{49} \text{ and }A=-\dfrac{145}{49} \end{array}$$
4.) Calculation of the coefficient $B$ from the initial value $x(0)=1$ :

$$\begin{array}{lll} x(t)&=-5y(t)-0.5y'(t)\\ &=-5Ae^{-8t}-5Bte^{-8t}-5Ce^{-t}+(-0.5B+4A)e^{-8t}+4Bte^{-8t}+0.5Ce^{-t}\\ &=(-A-0.5B)e^{-8t}-Bte^{-8t}-4.5Ce^{-t}\\ x(0)&=1=-A-0.5B-4.5C=\dfrac{145+9}{49}-\dfrac{B}{2} \text{ and }B=\dfrac{210}{49} \end{array}$$
5.) Summarizing the solution:

$$\begin{array}{lll} 49\begin{pmatrix}x(t)\\y(t)\end{pmatrix} &=\begin{pmatrix} 40 \\ -145\end{pmatrix}e^{-8t} +\begin{pmatrix} -210 \\ 210 \end{pmatrix}te^{-8t} +\begin{pmatrix} 9 \\ -2 \end{pmatrix}e^{-t}\\[16pt] 49\begin{pmatrix}x'(t)\\y'(t)\end{pmatrix} &=\begin{pmatrix} -530 \\ 1370 \end{pmatrix}e^{-8t} +\begin{pmatrix} 1680 \\ -1680 \end{pmatrix}te^{-8t} +\begin{pmatrix} -9 \\ 2 \end{pmatrix}e^{-t} \end{array}$$
6.) Confirmation of the correctness by checking the IVP:

$$\begin{array}{lll} &49\cdot\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} 40\\ -145 \end{pmatrix}e^{-8t} +\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} -210 \\ 210 \end{pmatrix}te^{-8t} +\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix} 9\\-2 \end{pmatrix}e^{-t}\\[16pt] &=\begin{pmatrix} 530\\-1370\end{pmatrix}e^{-8t} +\begin{pmatrix} -1680\\ 1680 \end{pmatrix}te^{-8t} +\begin{pmatrix} 58\\-2 \end{pmatrix}e^{-t} \end{array}$$
$$\begin{array}{lll} 49\begin{pmatrix}x'(t)\\y'(t)\end{pmatrix}+49\cdot\begin{pmatrix}6&-2\\2&10\end{pmatrix}\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=\begin{pmatrix} 49 \\ 0 \end{pmatrix}e^{-t} \end{array}$$

 

[mario99]

Guess, I should review the Laplace transformation calculus.

It is a must-learn tool. This powerful tool can save your mathematical life even in the toughest situations such as the Schrödinger Partial Differential Equation.

Baker-Campbell-Hausdorff formula to correct the difference $e^{X}e^{Y} - e^{X+Y}.$ It is the formal reason for what you called a "headache of thinking".

I have heard of it before but my math level does not have the capability to understand it.

My reviewed solution (and final confirmation of the correct result) is, for the sake of completeness:

By the way your final confirmation is a mathematical piece of Art. And it seems that you, I, and professor Dan are the only three guys here who are taking Differential Equations seriously. (At least this is what I think.)

 

[fresh_42]

And it seems that you, I, and professor Dan are the only three guys here who are taking Differential Equations seriously. (At least this is what I think.)

Differential equations are everywhere! If you are interested in a little non-technical, more entertaining article about differential equations you may want to read https://www.physicsforums.com/insights/differential-equation-systems-and-nature/