Solve the system of differential equations: dot x=-x+2y, dot…
Solve the system of differential equations: $$ \dot x=-x+2y,\qquad \dot y=-3x+4y+\frac{e^{3t}}{e^{2t}+1} $$
Explanation of Method
The forcing term $e^{3t}/(e^{2t}+1)$ does not match any exponential resonance pattern, so undetermined coefficients fails; variation of parameters using the fundamental matrix reduces the problem to two elementary substitution integrals (a logarithm and an arctangent).
- Find the eigenvalues and eigenvectors of $A$ to build the fundamental matrix $\Phi(t)$.
- Invert $\Phi(t)$.
- Compute $\Phi^{-1}(t)\mathbf f(t)$ and integrate each component.
- Form the particular solution $\mathbf u_p=\Phi(t)\int\Phi^{-1}\mathbf f\,dt$.
Step 1 · Eigenvalues and fundamental matrix
With $A=\begin{pmatrix}-1&2\\-3&4\end{pmatrix}$, $\det(A-\lambda I)=(\lambda-1)(\lambda-2)=0$, giving eigenvectors $(1,1)$ for $\lambda=1$ and $(2,3)$ for $\lambda=2$:
$$ \Phi(t)=\begin{pmatrix}e^{t}&2e^{2t}\\e^{t}&3e^{2t}\end{pmatrix},\qquad \Phi^{-1}(t)=\begin{pmatrix}3e^{-t}&-2e^{-t}\\-e^{-2t}&e^{-2t}\end{pmatrix}. $$
Step 2 · Compute $\Phi^{-1}\mathbf f$
$$ \Phi^{-1}(t)\mathbf f(t)=\begin{pmatrix}\dfrac{-2e^{2t}}{e^{2t}+1}\\[10pt]\dfrac{e^{t}}{e^{2t}+1}\end{pmatrix}. $$
Step 3 · Integrate each component
With $u=e^{2t}+1$ for the first and $u=e^t$ for the second,
$$ v_1=-\ln(e^{2t}+1),\qquad v_2=\arctan(e^{t}). $$
Step 4 · Particular solution
$$ \mathbf u_p=\Phi(t)\begin{pmatrix}v_1\\v_2\end{pmatrix}=\begin{pmatrix}-e^{t}\ln(e^{2t}+1)+2e^{2t}\arctan(e^{t})\\[4pt]-e^{t}\ln(e^{2t}+1)+3e^{2t}\arctan(e^{t})\end{pmatrix}. $$
Step 5 · Verification
Differentiating $x_p,y_p$ using the product rule and the derivatives of $\ln(e^{2t}+1)$ and $\arctan(e^t)$ confirms both original equations hold identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=1,\ \lambda_2=2$ |
| Integration substitutions | $u=e^{2t}+1$ (logarithm) and $u=e^t$ (arctangent) |
| Method | variation of parameters (undetermined coefficients does not apply) |
Additional Notes
- The forcing term is a rational function of $e^t$, not a pure exponential or polynomial, so undetermined coefficients has no matching ansatz — variation of parameters is the only systematic route.
- Both resulting integrals are elementary once the right substitution is spotted: $u=e^{2t}+1$ for the logarithm, $u=e^t$ for the arctangent.
- Verified by direct differentiation of the particular solution and substitution into both original equations.
Result
The general solution of the system is: