Solve the system of differential equations: dot x=x+2y, dot…
Solve the system of differential equations: $$ \dot x=x+2y,\qquad \dot y=x-5\sin t $$
Explanation of Method
Since $\pm i$ (the frequency of $\sin t$) is not an eigenvalue of $A$, a plain $\cos t,\sin t$ vector ansatz finds the particular solution by matching coefficients of $\cos t$ and $\sin t$ separately.
- Write the system in matrix form and find the eigenvalues of $A$.
- Find an eigenvector for each eigenvalue.
- Assemble the homogeneous solution.
- Try $\mathbf u_p=(a,b)\cos t+(c,d)\sin t$ and solve for the four coefficients.
Step 1 · Characteristic equation and eigenvectors
With $A=\begin{pmatrix}1&2\\1&0\end{pmatrix}$, $\det(A-\lambda I)=(\lambda-2)(\lambda+1)=0$, giving $\lambda_1=2$ with eigenvector $(2,1)$ and $\lambda_2=-1$ with eigenvector $(-1,1)$.
Step 2 · Homogeneous solution
$$ \mathbf u_h=C_1e^{2t}\begin{pmatrix}2\\1\end{pmatrix}+C_2e^{-t}\begin{pmatrix}-1\\1\end{pmatrix}. $$
Step 3 · Particular solution
Matching $\cos t$ and $\sin t$ coefficients after substituting $\mathbf u_p=(a,b)\cos t+(c,d)\sin t$ gives $a=-1,\ b=2,\ c=3,\ d=-1$:
$$ \mathbf u_p=\begin{pmatrix}-1\\2\end{pmatrix}\cos t+\begin{pmatrix}3\\-1\end{pmatrix}\sin t. $$
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 | $-5\sin t$ in the $y$-equation only |
| Particular solution | $(-\cos t+3\sin t,\ 2\cos t-\sin t)$ |
Additional Notes
- Even though the forcing only appears in the second equation, the coupling in $A$ means both $x_p$ and $y_p$ pick up nonzero $\cos t$ and $\sin t$ terms.
- Since neither eigenvalue is $\pm i$, no resonance correction (extra factor of $t$) is needed for the trigonometric ansatz.
- Verified by direct substitution into both original equations.
Result
The general solution of the system is: