Solving a system of differential equations by fundamental 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!

 

[BvU]

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

\begin {matrix}
t^2*(2 & 3) \\
(-2 & -3) 
\end{matrix}

But if you want brackets you can either do it yourself (outside the matrix)

 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

 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 ?

$\$

 

[arhzz]

You were close with

\begin {matrix}
t^2*(2 & 3) \\
(-2 & -3)
\end{matrix}

But if you want brackets you can either do it yourself (outside the matrix)

 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

 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?

 

[pasmith]

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.

 

[arhzz]

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

 

[pasmith]

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

 

[Mayhem]

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.

 

[vela]

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 $x_2$.

 

[arhzz]

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!