Solve the system of differential equations: (dx)/(dt) = x +…
Solve the system of differential equations: $$ \frac{dx}{dt} = x + 2y + t,\qquad \frac{dy}{dt} = 2x + y + t $$
Explanation of Method
The coefficient matrix is symmetric with two distinct real eigenvalues, so the homogeneous solution is a simple combination of two exponential modes; a linear ansatz handles the linear forcing.
- Write the system in matrix form and find the eigenvalues.
- Find an eigenvector for each eigenvalue.
- Build the homogeneous solution.
- Find a particular solution linear in $t$.
- Combine for the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$,
$$ \det(A-\lambda I)=(1-\lambda)^2-4=(\lambda-3)(\lambda+1)=0, $$
giving $\lambda_1=3$ and $\lambda_2=-1$.
Step 2 · Eigenvectors
$$ \mathbf v_1=\begin{pmatrix}1\\1\end{pmatrix}\ (\lambda_1=3),\qquad \mathbf v_2=\begin{pmatrix}1\\-1\end{pmatrix}\ (\lambda_2=-1). $$
Step 3 · Homogeneous solution
$$ \begin{pmatrix}x_h\\y_h\end{pmatrix}=C_1e^{3t}\begin{pmatrix}1\\1\end{pmatrix}+C_2e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}. $$
Step 4 · Particular solution
Trying $\mathbf u_p=(at+b,\,ct+d)^{\mathsf T}$ and matching coefficients gives $a=c=-\tfrac13$, $b=d=-\tfrac19$, so
$$ \mathbf u_p=\begin{pmatrix}-\tfrac13 t-\tfrac19\\[4pt]-\tfrac13 t-\tfrac19\end{pmatrix}. $$
Step 5 · General solution
$$ x(t)=C_1e^{3t}+C_2e^{-t}-\tfrac13 t-\tfrac19,\qquad y(t)=C_1e^{3t}-C_2e^{-t}-\tfrac13 t-\tfrac19. $$
Step 6 · Verification
Substituting $x(t)$ and $y(t)$ into $x'=x+2y+t$ and $y'=2x+y+t$ reproduces both derivatives exactly. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_1=3,\ \lambda_2=-1$ |
| Eigenvectors | $(1,1)$ and $(1,-1)$ |
| Particular solution | $-\tfrac13 t-\tfrac19$ in both components |
Additional Notes
- Because $A$ is symmetric, its eigenvalues are guaranteed real and its eigenvectors orthogonal.
- The forcing vector $(t,t)^{\mathsf T}$ is not itself an eigenvector direction, but the linear ansatz still works since $0$ is not an eigenvalue.
- Checked by back-substitution into both original equations.
Result
The general solution of the system is: