Solve the system of differential equations: (dx_1)/(dt) =…
Solve the system of differential equations: $$ \frac{dx_1}{dt} = x_1 - x_2,\qquad \frac{dx_2}{dt} = x_1 + 3x_2 $$
Explanation of Method
As in Problem 37, the coefficient matrix has a repeated eigenvalue with only a one-dimensional eigenspace, so a generalised eigenvector is again 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 $A=\begin{pmatrix}1&-1\\1&3\end{pmatrix}$,
$$ \det(A-\lambda I)=(1-\lambda)(3-\lambda)+1=(\lambda-2)^2=0, $$
a repeated eigenvalue $\lambda=2$.
Step 2 · Eigenvector and generalised eigenvector
Solving $(A-2I)\mathbf v=0$ gives $\mathbf v_1=(1,-1)$ (rank of $A-2I$ is 1, so a generalised eigenvector is needed). Solving $(A-2I)\mathbf v_2=\mathbf v_1$ gives $\mathbf v_2=(-1,0)$.
Step 3 · General solution
$$ x_1(t)=C_1e^{2t}+C_2(t-1)e^{2t},\qquad x_2(t)=-C_1e^{2t}-C_2\,t\,e^{2t}. $$
Step 4 · Verification
Differentiating and substituting confirms $\dot x_1=x_1-x_2$ and $\dot x_2=x_1+3x_2$ hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalue | $\lambda=2$ (multiplicity 2) |
| Eigenvector | $(1,-1)$ |
| Generalised eigenvector | $(-1,0)$ |
Additional Notes
- As in Problem 37, the geometric multiplicity of the repeated eigenvalue is only 1, so the chain $\mathbf v_1\to\mathbf v_2$ via $(A-2I)\mathbf v_2=\mathbf v_1$ is essential.
- Both components grow like $t\,e^{2t}$ for large $t$, with the polynomial factor $t$ eventually dominating the plain exponential term.
- Verified by direct substitution into both original equations.
Result
The general solution of the system is: