Solve the initial value problem: dot x=-4(x+y), dot…
Solve the initial value problem: $$ \dot x=-4(x+y),\qquad \dot y=-y-\tfrac14\dot x,\qquad x(0)=1,\ y(0)=0 $$
Explanation of Method
Substituting $\dot x=-4(x+y)$ into the second equation eliminates $\dot x$ and reduces the system to standard form $\dot y=x$; the resulting matrix has a repeated eigenvalue needing a generalised eigenvector, and the initial data then fixes both constants.
- Substitute $\dot x=-4(x+y)$ into the second equation to eliminate $\dot x$.
- Find the repeated eigenvalue of the resulting standard-form matrix and its Jordan chain.
- Assemble the general solution.
- Apply the initial conditions and verify.
Step 1 · Reduce to standard form
Substituting $\dot x=-4(x+y)$ into $\dot y=-y-\tfrac14\dot x$ gives $\dot y=-y+(x+y)=x$, so the system becomes
$$ \dot x=-4x-4y,\qquad \dot y=x. $$
Step 2 · Repeated eigenvalue and Jordan chain
With $A=\begin{pmatrix}-4&-4\\1&0\end{pmatrix}$, $\det(A-\lambda I)=(\lambda+2)^2=0$, a repeated eigenvalue $\lambda=-2$ with eigenvector $\mathbf v_1=(-2,1)$ and generalised eigenvector $\mathbf v_2=(1,0)$.
Step 3 · General solution
$$ x(t)=e^{-2t}\bigl(-2C_1+C_2(1-2t)\bigr),\qquad y(t)=e^{-2t}(C_1+C_2t). $$
Step 4 · Apply the initial conditions
From $x(0)=1,\ y(0)=0$: $C_1=0$ and $C_2=1$.
Step 5 · Verification
$x(0)=1$, $y(0)=0$ check out, and substituting confirms both the reduced system and the original equation $\dot y=-y-\tfrac14\dot x$ hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Reduced system eigenvalue | $\lambda=-2$ (repeated, needs generalised eigenvector) |
| Constants from ICs | $C_1=0,\ C_2=1$ |
| Long-term behaviour | $x,y\to0$ as $t\to\infty$ (stable improper node) |
Additional Notes
- The original system is not in standard form since $\dot x$ appears on the right-hand side of the second equation; eliminating it algebraically is the essential first step before any eigenvalue analysis.
- With $C_1=0$, the solution reduces to the pure generalised-eigenvector mode $C_2\,e^{-2t}(1-2t,t)$, without any contribution from the plain eigenvector mode.
- Verified against both the reduced standard-form system and the original (unreduced) second equation.
Result
The particular solution satisfying the initial data is: