Solve the system of differential equations: dot x=-3x+4y-2z…
Solve the system of differential equations: $$ \dot x=-3x+4y-2z,\qquad \dot y=x+z,\qquad \dot z=6x-6y+5z $$
Explanation of Method
As in the previous problem, three distinct real eigenvalues make the matrix diagonalisable; the eigenvectors here happen to have a particularly simple structure (each has one zero entry), which makes the final solution easy to read off.
- Write the system in matrix form and compute the characteristic polynomial.
- Solve the cubic for the eigenvalues.
- Find an eigenvector for each eigenvalue.
- Assemble and verify the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}-3&4&-2\\1&0&1\\6&-6&5\end{pmatrix}$, expanding $\det(A-\lambda I)$ gives
$$ -\lambda^3+2\lambda^2+\lambda-2=-(\lambda-1)(\lambda-2)(\lambda+1)=0. $$
Three distinct real eigenvalues: $\lambda_1=-1,\ \lambda_2=1,\ \lambda_3=2$.
Step 2 · Eigenvectors
$$ \mathbf v_1=\begin{pmatrix}-1\\0\\1\end{pmatrix}\ (\lambda=-1),\qquad \mathbf v_2=\begin{pmatrix}1\\1\\0\end{pmatrix}\ (\lambda=1),\qquad \mathbf v_3=\begin{pmatrix}0\\1\\2\end{pmatrix}\ (\lambda=2). $$
Each is verified directly against $A\mathbf v=\lambda\mathbf v$. ✓
Step 3 · Linear independence and general solution
The determinant of $[\mathbf v_1\ \mathbf v_2\ \mathbf v_3]$ is $-1\neq0$, so the three exponential modes are independent.
Step 4 · Verification
Direct substitution into all three original equations, collecting coefficients of $e^{-t}$, $e^{t}$, $e^{2t}$, confirms every equation holds. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=-1,\ \lambda_2=1,\ \lambda_3=2$ |
| Eigenvectors | $(-1,0,1),\ (1,1,0),\ (0,1,2)$ |
| Structure | diagonalisable |
Additional Notes
- Note that $z$ carries no $e^t$ contribution and $x$ carries no $e^{2t}$ contribution — a direct consequence of the zero entries in $\mathbf v_2$ and $\mathbf v_3$.
- Verified by substitution into all three original equations.
Result
The general solution of the system is: