Solve the system of differential equations: dot x=x+y+1+e^t…
Solve the system of differential equations: $$ \dot x=x+y+1+e^t,\qquad \dot y=3x-y $$
Explanation of Method
The forcing splits into a constant piece and an $e^t$ piece; since neither $\mu=0$ nor $\mu=1$ is an eigenvalue of $A$, each piece is matched separately with its own simple algebraic solve.
- Write the system in matrix form and find the eigenvalues of $A$.
- Find an eigenvector for each eigenvalue.
- Assemble the homogeneous solution.
- Solve separately for the constant part and the $e^t$ part of the particular solution.
Step 1 · Characteristic equation and eigenvectors
With $A=\begin{pmatrix}1&1\\3&-1\end{pmatrix}$, $\det(A-\lambda I)=\lambda^2-4=0$, giving $\lambda_1=2$ with eigenvector $(1,1)$ and $\lambda_2=-2$ with eigenvector $(1,-3)$.
Step 2 · Homogeneous solution
$$ \mathbf u_h=C_1e^{2t}\begin{pmatrix}1\\1\end{pmatrix}+C_2e^{-2t}\begin{pmatrix}1\\-3\end{pmatrix}. $$
Step 3 · Particular solution
For the constant piece, $A(a,b)^\top=(-1,0)$ gives $(a,b)=(-\tfrac14,-\tfrac34)$. For the $e^t$ piece, $(A-I)(c,d)^\top=(-1,0)$ gives $(c,d)=(-\tfrac23,-1)$:
$$ \mathbf u_p=\begin{pmatrix}-\tfrac14\\-\tfrac34\end{pmatrix}+e^t\begin{pmatrix}-\tfrac23\\-1\end{pmatrix}. $$
Step 4 · Verification
Substituting the general solution into both original equations confirms each holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=2,\ \lambda_2=-2$ |
| Constant part of $\mathbf u_p$ | $(-\tfrac14,-\tfrac34)$ |
| $e^t$ part of $\mathbf u_p$ | $(-\tfrac23,-1)$ |
Additional Notes
- Splitting the forcing $1+e^t$ into a constant piece and an exponential piece and solving each independently is valid by linearity, and avoids having to guess a combined ansatz.
- Since $\det A=-4\neq0$ and $\mu=1$ is not an eigenvalue, both algebraic solves have a unique solution with no resonance term.
- Verified by direct substitution into both original equations.
Result
The general solution of the system is: