I have this problem to solve the diffrential equation, can someone give me a hint where I should start
$\displaystyle \left\{\begin{array}{cc}x'+5x-2y=e^t\\y'-x+6y=e^{-2t}\end{array}\right.$
This is easily solved by using methods from linear algebra. The system can be written as:
$\displaystyle \left(\begin{array}{c}x_1\\x_2\end{array}\right)'= \left(\begin{array}{cc}-5&2\\1&-6\end{array}\right)\left(\begin{array}{c}x_1\\x_2\ end{array}\right)+\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)$
The eigenvalues that belong to the transformation matrix are -4 and -7, so you end up with the eigenvectors:
$\displaystyle T=\left(\begin{array}{cc}2&1\\1&-1\end{array}\right)$
With inverse:
$\displaystyle T^{-1}=\frac{1}{3}\left(\begin{array}{cc}1&1\\1&-2\end{array}\right)$
The new transformation matrix becomes: $\displaystyle A_f=\left(\begin{array}{cc}-4&0\\0&-7\end{array}\right)$
And the system is transformed to:
$\displaystyle \left(\begin{array}{c}y_1\\y_2\end{array}\right)' = A_fY + T^{-1}\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)=\left(\begin{array}{cc}-4&0\\0&-7\end{array}\right)\left(\begin{array}{c}y_1\\y_2\ end{array}\right)+\frac{1}{3}\left(\begin{array}{c c}1&1\\1&-2\end{array}\right)\left(\begin{array}{c}e^t\\e^{-2t}\end{array}\right)$
Then just solve the two systems independently and transform the solution to the old coordinates by using: $\displaystyle X=TY$
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:
$\displaystyle x = y' + 6y - e^{-2t}$ .... (A)
$\displaystyle \Rightarrow x' = y'' + 6y' + 2 e^{-2t}$ .... (B)
Substitute (A) and (B) into the first equation and re-arrange:
$\displaystyle y'' + 11 y' + 28 y = 3 e^{-2t} + e^t$ .... (C)
Solve (C) for $\displaystyle y = y(t)$. Substitute the solution into (A) to get $\displaystyle x = x(t)$.