Solve the system of differential equations: (dx_1)/(dt) =…
Solve the system of differential equations: $$ \frac{dx_1}{dt} = 2x_1 - x_2,\qquad \frac{dx_2}{dt} = x_1 $$
Explanation of Method
The coefficient matrix has a repeated eigenvalue with only a one-dimensional eigenspace, so a generalised eigenvector is required to build the second independent solution.
- Write the system in matrix form and solve the characteristic equation.
- Find the eigenvector for the repeated eigenvalue.
- Find a generalised eigenvector.
- Combine into the general solution.
Step 1 · Characteristic equation
With $A=\begin{pmatrix}2&-1\\1&0\end{pmatrix}$,
$$ \det(A-\lambda I)=(2-\lambda)(-\lambda)+1=(\lambda-1)^2=0, $$
a repeated eigenvalue $\lambda=1$.
Step 2 · Eigenvector and generalised eigenvector
Solving $(A-I)\mathbf v=0$ gives $\mathbf v_1=(1,1)$; solving $(A-I)\mathbf v_2=\mathbf v_1$ gives $\mathbf v_2=(1,0)$.
Step 3 · General solution
$$ x_1(t)=C_1e^{t}+C_2(t+1)e^{t},\qquad x_2(t)=C_1e^{t}+C_2\,t\,e^{t}. $$
Step 4 · Verification
Differentiating and substituting confirms $\dot x_1=2x_1-x_2$ and $\dot x_2=x_1$ hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalue | $\lambda=1$ (multiplicity 2) |
| Eigenvector | $(1,1)$ |
| Generalised eigenvector | $(1,0)$ |
Additional Notes
- The geometric multiplicity of $\lambda=1$ is only 1 (rank of $A-I$ is 1), so a chain of generalised eigenvectors is essential here.
- The solution grows without bound as $t\to\infty$ due to the positive eigenvalue, with the $t\,e^t$ term dominating for large $t$.
- Verified by direct substitution into both original equations.
Result
The general solution of the system is: