Solve the system of differential equations: dot x=x-y+sec t…
Solve the system of differential equations: $$ \dot x=x-y+\sec t,\qquad \dot y=2x-y $$
Explanation of Method
The homogeneous part oscillates ($\lambda=\pm i$), and the secant forcing has no matching undetermined-coefficients ansatz; variation of parameters reduces the problem to integrating $1\mp\tan t$, both elementary.
- Find the (purely imaginary) eigenvalues and build the fundamental matrix from the real and imaginary parts of the complex eigenvector.
- Invert the fundamental matrix.
- Compute $\Phi^{-1}(t)\mathbf f(t)$, giving $1-\tan t$ and $1+\tan t$.
- Integrate both and assemble the particular solution.
Step 1 · Eigenvalues and fundamental matrix
With $A=\begin{pmatrix}1&-1\\2&-1\end{pmatrix}$, $\det(A-\lambda I)=\lambda^2+1=0$, giving $\lambda=\pm i$ with eigenvector $(1,1-i)=\mathbf a+i\mathbf b$, $\mathbf a=(1,1)$, $\mathbf b=(0,-1)$:
$$ \Phi(t)=\begin{pmatrix}\cos t&\sin t\\\cos t+\sin t&\sin t-\cos t\end{pmatrix},\qquad \det\Phi=-1. $$
Step 2 · Invert and compute $\Phi^{-1}\mathbf f$
With $\mathbf f(t)=(\sec t,0)^\top$,
$$ \Phi^{-1}(t)\mathbf f(t)=\begin{pmatrix}1-\tan t\\1+\tan t\end{pmatrix}. $$
Step 3 · Integrate
$$ v_1=\int(1-\tan t)\,dt=t+\ln|\cos t|,\qquad v_2=\int(1+\tan t)\,dt=t-\ln|\cos t|. $$
Step 4 · Particular solution
With $L=\ln|\cos t|$,
$$ x_p=t(\cos t+\sin t)+L(\cos t-\sin t),\qquad y_p=2t\sin t+2L\cos t. $$
Step 5 · Verification
Using $L'=-\tan t$, direct differentiation of $x_p$ and $y_p$ and substitution confirms $\dot x_p=x_p-y_p+\sec t$ and $\dot y_p=2x_p-y_p$ hold exactly. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda=\pm i$ |
| Reduced integrals | $\int(1\mp\tan t)\,dt$ |
| Particular solution | involves $t\cos t,\ t\sin t,\ \ln|\cos t|\cos t,\ \ln|\cos t|\sin t$ |
Additional Notes
- The secant forcing has no elementary undetermined-coefficients ansatz, but variation of parameters always applies as long as the homogeneous fundamental matrix is known — here it reduces to two elementary trig integrals.
- The particular solution is only valid where $\cos t\neq0$, i.e.\ away from $t=\tfrac\pi2+k\pi$, matching where $\sec t$ itself is undefined.
- Verified directly: differentiating $x_p$ and $y_p$ and substituting reproduces both original equations exactly (with $L'=-\tan t$ the key derivative identity used).
Result
The general solution of the system is: