# Solving a system of differential equations by fundamental matrix

• arhzz
In summary: I think I lost my y_1 somewhere along the way. I found the eigenvalues and eigenvectors like this;$$y_1 = (v1,v2) * e^{\lambda_1*t}$$ and analog with ##\lambda_2## and the eigenvectors of that eigenvalue My fundamental matrix looks like this \begin {matrix}1 & e^{-t}\\ \frac{-2}{3} & -e^{-t})\end{matrix}
arhzz
Homework Statement
Solve the differential equation using the fundemental matrix and derivation of the constant
Relevant Equations
-
I am given this system of differential equations;
$$x_1'=2t^2x_1+3t^2x_2+t^5$$
$$x_2' =-2t^2x_1-3t^2x_2 +t^2$$

Now the first question states the following;
Find a fundamental matrix of the corresponding homogeneous system and
explain exactly how you arrive at independent solutions

And the second question;

We consider the DGL system from Exercise 15. Find all the solutions
with variation of the constants.

Now really it is just like having a part a) and b) of question.To solve the first question (part a if you will) I did this;

So the corresponding homogenous system should be

$$x_1'=2t^2x_1+3t^2x_2$$
$$x_2' =-2t^2x_1-3t^2x_2$$

Than I created the coefficent matrix and find the eigenvalues and eigenvectors;\begin {matrix}
t^2*(2 & 3) \\
(-2 & -3)
\end{matrix}

Note: the ##t^2 ## should be infront of the matrix multiplying the whole thing,but not sure how to do it in LaTeX properly;
I will for the sake of the length of the post skip how I got the eigenvalues but they should be ## \lambda_1 = 0 ## and ##\lambda_2 = -t^2 ##

Now I found the eigenvectors using wolfram alpha and they should be ## v_1 = (1,\frac{-2}{3})^T ## and ## v_2 =(1,-1)^T ## So these are the vectors that I need for my fundamental matrix.Now I have to create the fundamental matrix.I did that like this;

$$y_1 = (v1,v2) * e^{\lambda_1*t}$$ and analog with ##\lambda_2 ## and the eigenvectors of that eigenvalue

My fundamental matrix looks like this
\begin {matrix}
1 & e^{-t}\\
\frac{-2}{3} & -e^{-t})
\end{matrix}

Now I thought this was right,and the tutor for the class pointed out this; "the constant t should also be included". I am not really sure what that means and I am pretty sure that I cannot answer the second question if this one is not right. Do I simply need to multiply this matrix with t?

Thanks!

arhzz said:
Note: the t2 should be in front of the matrix multiplying the whole thing,but not sure how to do it in LaTeX properly;
You were close with
Code:
\begin {matrix}
t^2*(2 & 3) \\
(-2 & -3)
\end{matrix}
But if you want brackets you can either do it yourself (outside the matrix)
Code:
 t^2 \left ( \begin {matrix}
2 & 3 \\
-2 & -3
\end{matrix}\right )
to get
$$t^2 \left ( \begin {matrix} 2 & 3 \\ -2 & -3 \end{matrix}\right )$$
or get the exact same using pmatrix ##\Bigl(## less typing than \left ( ... \right ) ##\Bigr)##
$$t^2 \begin {pmatrix} 2 & 3 \\ -2 & -3 \end{pmatrix}$$
I forgot how to do e.g. right-aligning in the matrix environment, but in array it goes like
Code:
 t^2 \left ( \begin {array}{r}
2 & 3 \\
-2 & -3
\end{array}\right )
$$t^2 \left ( \begin {array}{r} 2 & 3 \\ -2 & -3 \end{array}\right )$$

And: in general you can right-click on ##\TeX## typeset stuff to see how it's done. Try it here and here .

Now for your exercise (I simply follow your footsteps -- that way I also learn a thing or two )

You have $$X'\left( t \right) = AX\left( t \right) + f\left( t \right)$$ and worked on $$X'\left( t \right) = AX\left( t \right) \ .$$ where $$A = t^2 \left ( \begin {array}{r} 2 & 3 \\ -2 & -3 \end{array}\right )$$
You find eigenvalues and -vectors but there you lose me. What is your ##y_1## ?

The method you use works for a system of differential equations with constant coefficients. Was that your exercise 15 ?

##\ ##

BvU said:
You were close with
Code:
\begin {matrix}
t^2*(2 & 3) \\
(-2 & -3)
\end{matrix}
But if you want brackets you can either do it yourself (outside the matrix)
Code:
 t^2 \left ( \begin {matrix}
2 & 3 \\
-2 & -3
\end{matrix}\right )
to get
$$t^2 \left ( \begin {matrix} 2 & 3 \\ -2 & -3 \end{matrix}\right )$$
or get the exact same using pmatrix ##\Bigl(## less typing than \left ( ... \right ) ##\Bigr)##
$$t^2 \begin {pmatrix} 2 & 3 \\ -2 & -3 \end{pmatrix}$$
I forgot how to do e.g. right-aligning in the matrix environment, but in array it goes like
Code:
 t^2 \left ( \begin {array}{r}
2 & 3 \\
-2 & -3
\end{array}\right )
$$t^2 \left ( \begin {array}{r} 2 & 3 \\ -2 & -3 \end{array}\right )$$

And: in general you can right-click on ##\TeX## typeset stuff to see how it's done. Try it here and here .

Now for your exercise (I simply follow your footsteps -- that way I also learn a thing or two )

You have $$X'\left( t \right) = AX\left( t \right) + f\left( t \right)$$ and worked on $$X'\left( t \right) = AX\left( t \right) \ .$$ where $$A = t^2 \left ( \begin {array}{r} 2 & 3 \\ -2 & -3 \end{array}\right )$$
You find eigenvalues and -vectors but there you lose me. What is your ##y_1## ?

The method you use works for a system of differential equations with constant coefficients. Was that your exercise 15 ?

##\ ##
Yea I am still trying to get the hang of matrices and vectors in latex. Although I find it pretty straight forward most of the time,can get hard to do complex expressions especially with the lack of examples online(at least I can't seem to find any)

Now to the exercise. I will try to explain a bit better. After findining the eigenvalues and eigenvectors,I tried building the Fundamental Matrix. Now I am not sure how you say it in english but approach would be the most fitting word.My approach was, using information of the internet on how to build the fundamental matrix was multiply the eigenvectors with ## e^{\lambda t} ## . Now we use this approach because we have "simple" eigenvectors (their eigenvalues are of multiplicity one). I am sure you know what I am talking about, if we had multiplicy the approach would be diffrent (more complicated) and for 3 even more complicated. The y1 and y2 are simply the vectors of the fundamentalsystem of which I would build the fundamental matrix.

Now I have been told (you pointed it out as well) for a system with constant coefficents. Now this leads me to believe that my general approach of tackling the problem was false since the next topic we are doing is systems with constant coefficents. Honestly this was the only way I found of getting solutions to these kind of problems. In our lecture notes is literally nothing helpful and every college from class has the same approach. So basically what it comes down to is this;

Can I use this approach on this problem, if not then how?

BvU
You can solve $$\mathbf{x}' = p(t)A\mathbf{x} + \mathbf{f}(t)$$ by diagonalizing $A$: $$\mathbf{y}' = P\mathbf{x}' = p(t)PAP^{-1}P\mathbf{x} + P\mathbf{f}(t) = p(t)\Lambda \mathbf{y} + P\mathbf{f}(t).$$ If $\Lambda$ is diagonal then this splits into $$y_i' - p(t)\lambda_i y_i = g_i(y)$$ which can be solved by an integrating factor.

pasmith said:
You can solve $$\mathbf{x}' = p(t)A\mathbf{x} + \mathbf{f}(t)$$ by diagonalizing $A$: $$\mathbf{y}' = P\mathbf{x}' = p(t)PAP^{-1}P\mathbf{x} + P\mathbf{f}(t) = p(t)\Lambda \mathbf{y} + P\mathbf{f}(t).$$ If $\Lambda$ is diagonal then this splits into $$y_i' - p(t)\lambda_i y_i = g_i(y)$$ which can be solved by an integrating factor.
Uh that math notation looks kind of scary,and we have not covered this topic in class (diagonalizing and those symbols don't look familiar at all) I'd be willing to try but I think that my knowlde of the matter is lacking to give it a solid attempt

Alternatively, the change of variable $u = \frac13 t^3$ will leave you with an inhomogenous system with constant coefficients.

arhzz said:
Yea I am still trying to get the hang of matrices and vectors in latex. Although I find it pretty straight forward most of the time,can get hard to do complex expressions especially with the lack of examples online(at least I can't seem to find any)

Now to the exercise. I will try to explain a bit better. After findining the eigenvalues and eigenvectors,I tried building the Fundamental Matrix. Now I am not sure how you say it in english but approach would be the most fitting word.My approach was, using information of the internet on how to build the fundamental matrix was multiply the eigenvectors with ## e^{\lambda t} ## . Now we use this approach because we have "simple" eigenvectors (their eigenvalues are of multiplicity one). I am sure you know what I am talking about, if we had multiplicy the approach would be diffrent (more complicated) and for 3 even more complicated. The y1 and y2 are simply the vectors of the fundamentalsystem of which I would build the fundamental matrix.

Now I have been told (you pointed it out as well) for a system with constant coefficents. Now this leads me to believe that my general approach of tackling the problem was false since the next topic we are doing is systems with constant coefficents. Honestly this was the only way I found of getting solutions to these kind of problems. In our lecture notes is literally nothing helpful and every college from class has the same approach. So basically what it comes down to is this;

Can I use this approach on this problem, if not then how?
https://www.codecogs.com/eqnedit.php this tool allows you to write out your LaTeX code before submitting. It also has a lot of templates which allow you to easily insert matrices.

arhzz said:
Uh that math notation looks kind of scary,and we have not covered this topic in class (diagonalizing and those symbols don't look familiar at all) I'd be willing to try but I think that my knowlde of the matter is lacking to give it a solid attempt
It's not too bad. Here's the idea w/o the matrix notation. Suppose you define ##y = a x_1 + b x_2## where ##a## and ##b## are constants chosen so that ##y' = \lambda t^2 y## where ##\lambda## is a constant. Then
\begin{align*}
y' &= (a x_1 + b x_2)' \\
&= a(2 t^2 x_1 + 3 t^2 x_2) + b(-2 t^2 x_1 - 3 t^2 x_2) \\
&= t^2 (a-b) (2 x_1 + 3 x_2)
\end{align*} One perhaps-obvious choice from looking at the righthand side would be ##a=2## and ##b=3## because then you'd have
$$y' = (2 x_1 + 3 x_2)' = (-1) t^2 (2 x_1 + 3 x_2) = (-1) t^2 y.$$ It's not a coincidence that the constant ##-1## is one of the eigenvalues you found earlier.

The other choice is to simply let ##a=b=1##. Then you have
$$y' = (x_1 + x_2)' = (0) t^2 (x_1 + x_2) = 0 t^2 y.$$ Note that ##\lambda = 0## was the other eigenvalue you found.

Let ##y_1 = 2 x_1 + 3 x_2## and ##y_2 = x_1 + x_2##. Solve the (inhomogeneous) differential equations they satisfy. Once you know ##y_1## and ##y_2##, you can solve for ##x_1## and ##x_2##.

BvU
Okay so I have not been able to solve this problem,even with the great help I got here.We did it in class yesterday and I know how to solve it now.Long story short you make a substituon to get a problem with constant coefficients,solve it than undo the substitution and there you have it.The solution should be this;

\begin {array}{r}

e^{-t^2/3} & 3 \\

-e^{-t^2/3} & -2

\end{array}

Thanks for the help!

BvU

## 1. What is a system of differential equations?

A system of differential equations is a set of equations that describe the relationship between multiple variables and their rates of change. These equations involve derivatives and can be used to model various physical, biological, or economic systems.

## 2. What is a fundamental matrix?

A fundamental matrix is a matrix that contains the solutions to a system of linear differential equations. It is used to find the general solution to the system and can also be used to solve initial value problems.

## 3. How do you solve a system of differential equations using a fundamental matrix?

To solve a system of differential equations using a fundamental matrix, you first need to find the fundamental matrix by solving the system of equations. Then, you can use the fundamental matrix to find the general solution by multiplying it by a vector of arbitrary constants. Finally, you can use initial conditions to find the specific solution to the system.

## 4. What are the benefits of using a fundamental matrix to solve a system of differential equations?

Using a fundamental matrix allows for a more efficient and organized method of solving a system of differential equations. It also provides a way to find the general solution to the system, which can then be used to find specific solutions for different initial conditions.

## 5. Are there any limitations to using a fundamental matrix to solve a system of differential equations?

One limitation of using a fundamental matrix is that it can only be used for linear systems of differential equations. Additionally, finding the fundamental matrix can be a complex and time-consuming process for more complicated systems.

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