Solve the system of differential equations: dot…
Solve the system of differential equations: $$ \dot x=3x-4y+e^{-2t},\qquad \dot y=x-2y-3e^{-2t} $$
Explanation of Method
The forcing frequency $\mu=-2$ does not coincide with either eigenvalue of $A$, so a plain exponential vector ansatz $e^{-2t}(a,b)$ finds the particular solution directly, without any resonance correction.
- Write the system in matrix form and find the eigenvalues of $A$.
- Find an eigenvector for each eigenvalue.
- Assemble the homogeneous solution.
- Since the forcing frequency is not an eigenvalue, solve $(A+2I)(a,b)^\top=-\mathbf c$ for the particular solution.
Step 1 · Characteristic equation and eigenvectors
With $A=\begin{pmatrix}3&-4\\1&-2\end{pmatrix}$, $\det(A-\lambda I)=(\lambda-2)(\lambda+1)=0$, giving $\lambda_1=2$ with eigenvector $(4,1)$ and $\lambda_2=-1$ with eigenvector $(1,1)$.
Step 2 · Homogeneous solution
$$ \mathbf u_h=C_1e^{2t}\begin{pmatrix}4\\1\end{pmatrix}+C_2e^{-t}\begin{pmatrix}1\\1\end{pmatrix}. $$
Step 3 · Particular solution
Since $\mu=-2$ is not an eigenvalue, try $\mathbf u_p=e^{-2t}(a,b)$. Substituting gives $(A+2I)(a,b)^\top=(-1,3)$, solved by
$$ a=3,\qquad b=4\ \Longrightarrow\ \mathbf u_p=e^{-2t}\begin{pmatrix}3\\4\end{pmatrix}. $$
Step 4 · Verification
Substituting the general solution into both original equations confirms each holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=2,\ \lambda_2=-1$ |
| Forcing frequency | $\mu=-2$ (not an eigenvalue — no resonance) |
| Particular solution | $e^{-2t}(3,4)$ |
Additional Notes
- Because none of the three frequencies involved ($2,-1,-2$) coincide, this is the simplest possible case of undetermined coefficients for a linear system — a single algebraic solve for $(a,b)$.
- As $t\to\infty$, the $e^{2t}$ homogeneous mode dominates unless $C_1=0$.
- Verified by direct substitution into both original equations.
Result
The general solution of the system is: