Solve the initial value problem: dot x=-x+y+z, dot y=x-y+z…
Solve the initial value problem: $$ \dot x=-x+y+z,\qquad \dot y=x-y+z,\qquad \dot z=x+y+z,\qquad x(0)=1,\ y(0)=0,\ z(0)=0 $$
Explanation of Method
Three distinct real eigenvalues make this a straightforward eigenvector decomposition; the zero initial conditions on $y$ and $z$ then pin down all three constants through a small linear system.
- Write the matrix $A$ and factor the characteristic cubic.
- Find an eigenvector for each of the three distinct eigenvalues.
- Assemble the general solution and apply the initial conditions.
- Solve the resulting linear system for the constants.
Step 1 · Characteristic equation
With $A=\begin{pmatrix}-1&1&1\\1&-1&1\\1&1&1\end{pmatrix}$, expanding and factoring gives
$$ -(\lambda+1)(\lambda-2)(\lambda+2)=0, $$
so $\lambda_1=-1,\ \lambda_2=2,\ \lambda_3=-2$.
Step 2 · Eigenvectors
$$ \mathbf v_1=\begin{pmatrix}1\\1\\-1\end{pmatrix}\ (\lambda_1=-1),\qquad \mathbf v_2=\begin{pmatrix}1\\1\\2\end{pmatrix}\ (\lambda_2=2),\qquad \mathbf v_3=\begin{pmatrix}-1\\1\\0\end{pmatrix}\ (\lambda_3=-2). $$
Step 3 · Apply the initial conditions
With $x(0)=1,\ y(0)=0,\ z(0)=0$, the resulting $3\times3$ linear system for $C_1,C_2,C_3$ solves to
$$ C_1=\frac{1}{3},\qquad C_2=\frac{1}{6},\qquad C_3=-\frac{1}{2}. $$
Step 4 · Verification
Substituting back confirms $x(0)=1$, $y(0)=0$, $z(0)=0$, and all three differential equations hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=-1,\ \lambda_2=2,\ \lambda_3=-2$ |
| Constants from ICs | $C_1=\tfrac13,\ C_2=\tfrac16,\ C_3=-\tfrac12$ |
| Long-term behaviour | dominated by $e^{2t}$ as $t\to\infty$ |
Additional Notes
- With three distinct real eigenvalues, applying initial conditions is a routine $3\times3$ linear solve — no resonance or degeneracy complicates the problem.
- Since $\lambda_2=2$ is the largest eigenvalue, the $e^{2t}$ term eventually dominates all three components for large $t$.
- Verified against the given initial data and by direct substitution into all three original equations.
Result
The particular solution satisfying the initial data is: