1. ## diffrential equations

I have this problem to solve the diffrential equation, can someone give me a hint where I should start
$\left\{\begin{array}{cc}x'+5x-2y=e^t\\y'-x+6y=e^{-2t}\end{array}\right.$

2. This is easily solved by using methods from linear algebra. The system can be written as:

$\left(\begin{array}{c}x_1\\x_2\end{array}\right)'= \left(\begin{array}{cc}-5&2\\1&-6\end{array}\right)\left(\begin{array}{c}x_1\\x_2\ end{array}\right)+\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)$

The eigenvalues that belong to the transformation matrix are -4 and -7, so you end up with the eigenvectors:

$T=\left(\begin{array}{cc}2&1\\1&-1\end{array}\right)$

With inverse:

$T^{-1}=\frac{1}{3}\left(\begin{array}{cc}1&1\\1&-2\end{array}\right)$

The new transformation matrix becomes: $A_f=\left(\begin{array}{cc}-4&0\\0&-7\end{array}\right)$

And the system is transformed to:

$\left(\begin{array}{c}y_1\\y_2\end{array}\right)' = A_fY + T^{-1}\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)=\left(\begin{array}{cc}-4&0\\0&-7\end{array}\right)\left(\begin{array}{c}y_1\\y_2\ end{array}\right)+\frac{1}{3}\left(\begin{array}{c c}1&1\\1&-2\end{array}\right)\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)$

Then just solve the two systems independently and transform the solution to the old coordinates by using: $X=TY$

3. I have only done Linear Algebra 103!

I don't understand this, sorry I wasn't clear enough about my math skills
This is in a chapter called "simultaneous differential equations".
So I can not return the solutions in linear algebric form, my teacher would defenitly see that I got some advanced help

4. Originally Posted by hlolli
I have only done Linear Algebra 103!

I don't understand this, sorry I wasn't clear enough about my math skills
This is in a chapter called "simultaneous differential equations".
So I can not return the solutions in linear algebric form, my teacher would defenitly see that I got some advanced help
In that case, from the second equation:

$x = y' + 6y - e^{-2t}$ .... (A)

$\Rightarrow x' = y'' + 6y' + 2 e^{-2t}$ .... (B)

Substitute (A) and (B) into the first equation and re-arrange:

$y'' + 11 y' + 28 y = 3 e^{-2t} + e^t$ .... (C)

Solve (C) for $y = y(t)$. Substitute the solution into (A) to get $x = x(t)$.

5. Originally Posted by mr fantastic
In that case, from the second equation:

$x = y' + 6y - e^{-2t}$ .... (A)

$\Rightarrow x' = y'' + 6y' + 2 e^{-2t}$ .... (B)

Substitute (A) and (B) into the first equation and re-arrange:

$y'' + 11 y' + 28 y = 3 e^{-2t} + e^t$ .... (C)

Solve (C) for $y = y(t)$. Substitute the solution into (A) to get $x = x(t)$.
Thank you, after a few hours I figured it out and I got the same. Now I know I did right!