0
$\begingroup$

Find the general solution for: $$y_1'=-y_1+3y_2 $$ $$y_2'=2y_1-2y_2 $$

Find the solution satisfying the initial condition $y(0)=\begin{pmatrix} 5 \\ 0 \\ \end{pmatrix}$

From the equation above, $\lambda_1=1$ and $\lambda_2=-4$,

The general soltution is: $$c_1e^x\begin{pmatrix} 3 \\ 2 \\ \end{pmatrix} + c_2e^{-4x}\begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}$$

I'm not sure how to use the condition above to find the particular solution.

$\endgroup$
2
$\begingroup$

Hint: solve the linear system $$c_1e^0\begin{pmatrix} 3 \\ 2 \\ \end{pmatrix} + c_2e^0\begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}=y(0)=\begin{pmatrix} 5 \\ 0 \end{pmatrix},$$ that is $$\begin{cases} 3c_1+c_2=5\\ 2c_1-c_2=0 \end{cases},$$ for the constants $c_1,c_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.