Solve the system of differential equations: dot…
Solve the system of differential equations: $$ \dot x=-4x-2y+\frac{2}{e^t-1},\qquad \dot y=6x+3y-\frac{3}{e^t-1} $$
Explanation of Method
One eigenvalue of $A$ is exactly zero, so the homogeneous solution includes a constant mode; the singular forcing $1/(e^t-1)$ again rules out undetermined coefficients, but variation of parameters reduces cleanly to a single logarithmic integral.
- Find the eigenvalues and eigenvectors of $A$ (noting one eigenvalue is zero).
- Build the fundamental matrix and invert it.
- Compute $\Phi^{-1}(t)\mathbf f(t)$; one component vanishes identically.
- Integrate the remaining component and assemble the particular solution.
Step 1 · Eigenvalues and fundamental matrix
With $A=\begin{pmatrix}-4&-2\\6&3\end{pmatrix}$, $\det(A-\lambda I)=\lambda(\lambda+1)=0$, giving eigenvectors $(1,-2)$ for $\lambda=0$ and $(2,-3)$ for $\lambda=-1$:
$$ \Phi(t)=\begin{pmatrix}1&2e^{-t}\\-2&-3e^{-t}\end{pmatrix},\qquad \Phi^{-1}(t)=\begin{pmatrix}-3&-2\\2e^{t}&e^{t}\end{pmatrix}. $$
Step 2 · Compute $\Phi^{-1}\mathbf f$
With $\mathbf f(t)=\dfrac{1}{e^t-1}(2,-3)^\top$,
$$ \Phi^{-1}(t)\mathbf f(t)=\begin{pmatrix}0\\[4pt]\dfrac{e^{t}}{e^{t}-1}\end{pmatrix}. $$
Step 3 · Integrate
The first component contributes nothing; with $u=e^t-1$,
$$ v_1=0,\qquad v_2=\ln(e^{t}-1). $$
Step 4 · Particular solution
$$ \mathbf u_p=\Phi(t)\begin{pmatrix}0\\\ln(e^t-1)\end{pmatrix}=\begin{pmatrix}2e^{-t}\ln(e^t-1)\\-3e^{-t}\ln(e^t-1)\end{pmatrix}. $$
Step 5 · Verification
Differentiating $x_p,y_p$ and substituting confirms both original equations hold identically, and the homogeneous ($\lambda=0$ and $\lambda=-1$) modes check out separately. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=0,\ \lambda_2=-1$ |
| Key simplification | one component of $\Phi^{-1}\mathbf f$ vanishes identically |
| Particular solution | $e^{-t}\ln(e^t-1)\cdot(2,-3)$ |
Additional Notes
- A zero eigenvalue means one homogeneous solution is simply constant in time — the system has a genuine equilibrium direction, along $(1,-2)$.
- The forcing vector happens to be proportional to the eigenvector direction of $\lambda=-1$ in a way that makes one component of $\Phi^{-1}\mathbf f$ vanish exactly, simplifying the integration to a single logarithm.
- Verified by direct differentiation and substitution into both original equations.
Result
The general solution of the system is: