Solve the initial value problem: dot x=3x+8y, dot y=-x-3y…

Solve the initial value problem: $$ \dot x=3x+8y,\qquad \dot y=-x-3y,\qquad x(0)=6,\ y(0)=-2 $$

Explanation of Method

Two distinct real eigenvalues make this a routine eigenvector decomposition; the given initial data then fixes both constants through a simple $2\times2$ linear solve.

Find the eigenvalues and eigenvectors of $A$.
Assemble the general solution.
Apply the initial conditions and solve for the constants.
Verify the particular solution.

Step 1 · Characteristic equation and eigenvectors

With $A=\begin{pmatrix}3&8\\-1&-3\end{pmatrix}$, $\det(A-\lambda I)=\lambda^2-1=0$, giving $\lambda_1=1$ with eigenvector $(-4,1)$ and $\lambda_2=-1$ with eigenvector $(-2,1)$.

Step 2 · Apply the initial conditions

From $x(0)=6,\ y(0)=-2$, solving the resulting $2\times2$ linear system gives $C_1=-1,\ C_2=-1$.

Step 3 · Verification

$x(0)=4+2=6$ and $y(0)=-1-1=-2$ check out, and substituting into both original equations confirms they hold identically. ✓

Summary Table

Quantity	Value
Eigenvalues	$\lambda_1=1,\ \lambda_2=-1$
Constants from ICs	$C_1=-1,\ C_2=-1$
Behaviour	$x,y\to\infty$ as $t\to\infty$ (dominated by $e^t$)

Additional Notes

With one positive and one negative eigenvalue, the origin is a saddle point; the given initial condition happens to place the trajectory on a combination of both modes rather than purely on the stable or unstable manifold.
Since $C_1,C_2\neq0$, both exponential modes are genuinely present, and the $e^t$ term eventually dominates for large $t$.
Verified against the given initial data and by direct substitution into both original equations.

Result

The particular solution satisfying the initial data is:

$$ \begin{aligned} x(t) &= 4e^{t}+2e^{-t},\\ y(t) &= -e^{t}-e^{-t}. \end{aligned} $$