Solve the system of differential equations: dot x=4x-3y+sin…
Solve the system of differential equations: $$ \dot x=4x-3y+\sin t,\qquad \dot y=2x-y-2\cos t $$
Explanation of Method
Two distinct real eigenvalues give a simple homogeneous solution; since $\mu=i$ is not an eigenvalue, a $\cos t,\sin t$ vector ansatz finds the particular solution by matching coefficients of $\cos t$ and $\sin t$ separately, which reduces to a small linear system for the four unknown coefficients.
- 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 $A=\begin{pmatrix}4&-3\\2&-1\end{pmatrix}$, $\det(A-\lambda I)=(\lambda-1)(\lambda-2)=0$, giving $\lambda_1=1$ with eigenvector $(1,1)$ and $\lambda_2=2$ with eigenvector $(3,2)$.
Step 2 · Homogeneous solution
$$ \mathbf u_h=C_1e^{t}\begin{pmatrix}1\\1\end{pmatrix}+C_2e^{2t}\begin{pmatrix}3\\2\end{pmatrix}. $$
Step 3 · Particular solution
Substituting $\mathbf u_p=(a,b)\cos t+(c,d)\sin t$ into the system and matching $\cos t$ and $\sin t$ coefficients gives a $4\times4$ linear system, solved by $a=1,\ b=2,\ c=-2,\ d=-2$:
$$ \mathbf u_p=\begin{pmatrix}1\\2\end{pmatrix}\cos t+\begin{pmatrix}-2\\-2\end{pmatrix}\sin t. $$
Step 4 · Verification
Substituting the general solution into both original equations confirms each holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=1,\ \lambda_2=2$ |
| Forcing | $\sin t$ in the $x$-equation, $-2\cos t$ in the $y$-equation |
| Particular solution | $(\cos t-2\sin t,\ 2\cos t-2\sin t)$ |
Additional Notes
- Because the forcing appears in both equations with different trigonometric functions ($\sin t$ vs.\ $\cos t$), the four undetermined coefficients $a,b,c,d$ must be solved simultaneously rather than component by component.
- Since neither eigenvalue is $\pm i$, no resonance correction (extra factor of $t$) is required for the trigonometric ansatz.
- Verified by direct substitution into both original equations.
Result
The general solution of the system is: