Solve the system of differential equations: dot…
Solve the system of differential equations: $$ \dot x=2x+y-2z-t+2,\qquad \dot y=1-x,\qquad \dot z=x+y-z-t+1 $$
Explanation of Method
The homogeneous part has one real eigenvalue and a purely imaginary pair, giving one exponential and two oscillatory modes; since the forcing is linear in $t$ and $t=0$ is not an eigenvalue, a linear-in-$t$ vector ansatz finds the particular solution directly.
- Write the system in matrix form $\dot{\mathbf u}=A\mathbf u+\mathbf f(t)$ and find the eigenvalues of $A$.
- Find the real eigenvector and the complex eigenvector, splitting the latter into real and imaginary parts.
- Assemble the homogeneous solution.
- Find a particular solution of the form $\mathbf p+\mathbf q\,t$.
- Combine for the general solution.
Step 1 · Characteristic equation
With $A=\begin{pmatrix}2&1&-2\\-1&0&0\\1&1&-1\end{pmatrix}$, expanding gives
$$ (1-\lambda)(1+\lambda^2)=0, $$
so $\lambda_1=1$ and $\lambda_{2,3}=\pm i$.
Step 2 · Eigenvectors
For $\lambda_1=1$: $\mathbf v_1=(1,-1,0)$. For $\lambda=i$: $\mathbf v=(1,i,1)=\mathbf a+i\mathbf b$ with $\mathbf a=(1,0,1)$, $\mathbf b=(0,1,0)$.
Step 3 · Homogeneous solution
$$ \mathbf u_h=C_1e^{t}\begin{pmatrix}1\\-1\\0\end{pmatrix}+C_2\begin{pmatrix}\cos t\\-\sin t\\\cos t\end{pmatrix}+C_3\begin{pmatrix}\sin t\\\cos t\\\sin t\end{pmatrix}. $$
Step 4 · Particular solution
Trying $\mathbf u_p=\mathbf p+\mathbf q\,t$ and matching the $t^1$ and $t^0$ coefficients of the forcing gives
$$ \mathbf q=\begin{pmatrix}0\\1\\0\end{pmatrix},\qquad \mathbf p=\begin{pmatrix}0\\0\\1\end{pmatrix}\ \Longrightarrow\ \mathbf u_p=\begin{pmatrix}0\\t\\1\end{pmatrix}. $$
Step 5 · Verification
Substituting the particular solution and each homogeneous mode separately into the original system confirms all three equations hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=1,\ \lambda_{2,3}=\pm i$ |
| Particular solution | $(x_p,y_p,z_p)=(0,t,1)$ |
| Structure | one exponential mode + one oscillatory pair |
Additional Notes
- Since $0$ is not an eigenvalue of $A$, the linear-in-$t$ ansatz for the particular solution is guaranteed to work without any resonance correction.
- The purely imaginary pair $\pm i$ means the oscillatory part of the homogeneous solution neither grows nor decays, while the $e^t$ mode dominates for large $t$.
- Verified by substituting both the particular solution and each homogeneous mode into the original system.
Result
The general solution of the system is: