Solve the system of differential equations: dot x=2x+y+2e^t…
Solve the system of differential equations: $$ \dot x=2x+y+2e^t,\qquad \dot y=x+2y-3e^{4t} $$
Explanation of Method
The forcing has two exponential pieces: $2e^t$ is resonant because $\mu=1$ coincides with an eigenvalue of $A$, requiring a $t\,e^t$ ansatz, while $-3e^{4t}$ is non-resonant and handled with a plain exponential ansatz.
- Write the system in matrix form and find the eigenvalues of $A$.
- Find an eigenvector for each eigenvalue.
- Solve the resonant piece ($\mu=1$) with a $t\,e^t+e^t$ ansatz, using the eigenvector direction for the $t\,e^t$ term.
- Solve the non-resonant piece ($\mu=4$) with a plain exponential ansatz.
- Combine and verify by direct substitution.
Step 1 · Characteristic equation and eigenvectors
With $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$, $\det(A-\lambda I)=(\lambda-1)(\lambda-3)=0$, giving $\lambda_1=1$ with eigenvector $\mathbf v_1=(-1,1)$ and $\lambda_2=3$ with eigenvector $(1,1)$.
Step 2 · Homogeneous solution
$$ \mathbf u_h=C_1e^{t}\begin{pmatrix}-1\\1\end{pmatrix}+C_2e^{3t}\begin{pmatrix}1\\1\end{pmatrix}. $$
Step 3 · Resonant piece ($\mu=1$)
Since $\mu=1=\lambda_1$, try $\mathbf u_p^{(1)}=t\,\mathbf a\,e^t+\mathbf b\,e^t$. Matching the $t\,e^t$ coefficient forces $\mathbf a=k\mathbf v_1$; the solvability (Fredholm) condition on the $e^t$ coefficient then fixes $k=-1$, giving $\mathbf a=(1,-1)$, and solving $(A-I)\mathbf b=\mathbf a-(2,0)$ gives
$$ \mathbf u_p^{(1)}=t\,e^t\begin{pmatrix}1\\-1\end{pmatrix}+e^t\begin{pmatrix}-1\\0\end{pmatrix}. $$
Step 4 · Non-resonant piece ($\mu=4$)
Since $\mu=4$ is not an eigenvalue, try $\mathbf u_p^{(2)}=e^{4t}(a,b)$. Solving $(A-4I)(a,b)^\top=(0,3)$ gives $a=-1,\ b=-2$:
$$ \mathbf u_p^{(2)}=e^{4t}\begin{pmatrix}-1\\-2\end{pmatrix}. $$
Step 5 · Verification
Substituting $\mathbf u=\mathbf u_h+\mathbf u_p^{(1)}+\mathbf u_p^{(2)}$ into both original equations and matching the $e^t,\ te^t,\ e^{3t},\ e^{4t}$ coefficients confirms every term cancels correctly. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=1,\ \lambda_2=3$ |
| Resonant forcing | $2e^t$ (since $\mu=1=\lambda_1$) → needs $t\,e^t$ term |
| Non-resonant forcing | $-3e^{4t}$ → plain exponential ansatz |
Additional Notes
- The resonant case is the delicate part: the coefficient of $t\,e^t$ must lie along the eigenvector direction, and a solvability condition (here forcing $k=-1$) is required before the $e^t$-only part can be found consistently.
- A common slip in this type of problem is choosing the $e^t$-only correction term inconsistently with the eigenvector direction, which silently breaks the equations by a constant multiple of $e^t$ — always double-check the resonant piece by direct substitution.
- Verified by substituting the complete general solution into both original equations and matching all four exponential/polynomial types.
Result
The general solution of the system is: