Solve the system of differential equations: dot…
Solve the system of differential equations: $$ \dot x=6x-72y+44z,\qquad \dot y=4x-43y+26z,\qquad \dot z=6x-63y+38z $$
Explanation of Method
Despite the large coefficients, the characteristic cubic factors cleanly into three distinct integer eigenvalues, one of which is zero; the corresponding eigenvector lies in the null space of $A$ and contributes a constant equilibrium term.
- Write the system in matrix form and compute the characteristic polynomial.
- Solve the cubic for the three eigenvalues.
- Find an eigenvector for each eigenvalue.
- Assemble and verify the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}6&-72&44\\4&-43&26\\6&-63&38\end{pmatrix}$, expanding $\det(A-\lambda I)$ gives
$$ -\lambda^3+\lambda^2+2\lambda=-\lambda(\lambda-2)(\lambda+1)=0. $$
Three distinct real eigenvalues: $\lambda_1=0,\ \lambda_2=2,\ \lambda_3=-1$.
Step 2 · Eigenvectors
$$ \mathbf v_1=\begin{pmatrix}2\\2\\3\end{pmatrix}\ (\lambda=0),\qquad \mathbf v_2=\begin{pmatrix}3\\2\\3\end{pmatrix}\ (\lambda=2),\qquad \mathbf v_3=\begin{pmatrix}4\\1\\1\end{pmatrix}\ (\lambda=-1). $$
Each is verified directly: $A\mathbf v_1=\mathbf 0$, $A\mathbf v_2=2\mathbf v_2$, $A\mathbf v_3=-\mathbf v_3$. ✓
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 modes are independent.
Step 4 · Verification
Substituting the general solution into all three equations and matching the constant, $e^{2t}$, and $e^{-t}$ coefficients confirms every equation holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=0,\ \lambda_2=2,\ \lambda_3=-1$ |
| Null-space vector | $(2,2,3)$ |
| Structure | diagonalisable, no Jordan blocks |
Additional Notes
- The $\lambda=0$ mode contributes a constant equilibrium vector $C_1(2,2,3)^\top$, lying in the null space of $A$.
- Note the elegant structure: $x=y$ along the constant mode, $x=z$ along the $\lambda_2$ mode, and $y=z$ along the $\lambda_3$ mode.
- Verified by direct substitution into all three equations.
Result
The general solution of the system is: