MHB -307.28.1 Find the general solution to the system of DE

Click For Summary
The discussion focuses on finding the general solution to a system of differential equations defined by y'_1 = y_1 + 5y_2 and y'_2 = -2y_1 - y_2. Participants analyze the characteristic equation derived from the coefficient matrix A, which is given as A = [[1, 5], [-2, -1]]. The determinant leads to the characteristic polynomial λ^2 + 9 = 0, indicating complex eigenvalues. The next step involves determining the values of λ, which are ±3i, leading to a solution involving exponential functions and trigonometric identities. The conversation emphasizes the methodical approach to solving the system of differential equations.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\tiny{28.1}$
2000
Find the general solution to the system of differential equations
\begin{align*}\displaystyle
y'_1&=y_1+5y_2\\
y'_2&=-2y_1+-y_2
\end{align*}
why is there a $+-y_2$ in the given
ok going to take this a step at a time... so..$A=\left[\begin{array}{c}1 & 5 \\ -2 & -1 \end{array}\right]$
then
$\left[\begin{array}{c}1-\lambda & 5 \\ -2 & -1-\lambda \end{array}\right]
=\lambda^2+9$ ?
 
Last edited:
Physics news on Phys.org
yes, [math]\left|\begin{array}{cc}1- \lambda & 5 \\ -2 & -1- \lambda\end{array}\right|= (1- \lambda)(-1- \lambda)+ 10= -1- \lambda+ \lambda+ \lambda^2+ 10= \lambda^2+ 9= 0[/math].

Now, what are the values of [math]\lambda[/math]?
 
Last edited by a moderator:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K