Solve the system of differential equations: dot…

Solve the system of differential equations: $$ \dot x_1=4x_1-x_2,\qquad \dot x_2=3x_1+x_2-x_3,\qquad \dot x_3=x_1+x_3 $$

Explanation of Method

The characteristic polynomial collapses to $-(\lambda-2)^3$, a single triple eigenvalue whose eigenspace is only one-dimensional; building the full solution requires a length-3 Jordan chain of one eigenvector and two generalised eigenvectors.

Compute the characteristic polynomial and confirm it is a perfect cube.
Find the (unique) eigenvector.
Solve for two successive generalised eigenvectors to complete the Jordan chain.
Assemble the three solutions using the standard Jordan-chain formula.

Step 1 · Characteristic equation

With $A=\begin{pmatrix}4&-1&0\\3&1&-1\\1&0&1\end{pmatrix}$, expanding gives

$$ \det(A-\lambda I)=-(\lambda-2)^3=0, $$

a triple eigenvalue $\lambda=2$.

Step 2 · Eigenvector

$A-2I$ has rank 2, so the geometric multiplicity is only 1. Solving $(A-2I)\mathbf v_1=0$ gives

$$ \mathbf v_1=\begin{pmatrix}1\\2\\1\end{pmatrix}. $$

Step 3 · Generalised eigenvectors

Solving $(A-2I)\mathbf v_2=\mathbf v_1$ gives $\mathbf v_2=(1,1,0)$; solving $(A-2I)\mathbf v_3=\mathbf v_2$ gives $\mathbf v_3=(0,-1,0)$, completing the Jordan chain.

Step 4 · Three solutions from the Jordan chain

$$ \mathbf x^{(1)}=e^{2t}\mathbf v_1,\quad \mathbf x^{(2)}=e^{2t}(t\mathbf v_1+\mathbf v_2),\quad \mathbf x^{(3)}=e^{2t}\!\left(\tfrac{t^2}{2}\mathbf v_1+t\mathbf v_2+\mathbf v_3\right). $$

Step 5 · Verification

Differentiating the resulting general solution and substituting confirms all three original equations hold identically. ✓

Summary Table

Quantity	Value
Eigenvalue	$\lambda=2$ (algebraic multiplicity 3, geometric multiplicity 1)
Jordan chain	$\mathbf v_1=(1,2,1)\to\mathbf v_2=(1,1,0)\to\mathbf v_3=(0,-1,0)$
Structure	single $3\times3$ Jordan block

Additional Notes

With a single Jordan block of size 3, every component of the solution grows like $t^2e^{2t}$ for large $t$, the strongest possible polynomial correction for a $3\times3$ system.
Each generalised eigenvector in the chain is only unique up to adding a multiple of the previous vector in the chain — the choices made here are simply the most convenient.
Verified by direct substitution of the explicit solution into all three original equations.

Result

The general solution of the system is:

$$ \begin{aligned} x_1(t) &= e^{2t}\!\left[C_1 + C_2(t+1) + C_3\!\left(\tfrac{t^2}{2}+t\right)\right],\\ x_2(t) &= e^{2t}\!\left[2C_1 + C_2(2t+1) + C_3(t^2+t-1)\right],\\ x_3(t) &= e^{2t}\!\left[C_1 + C_2\, t + C_3\,\tfrac{t^2}{2}\right]. \end{aligned} $$