Solve the system of differential equations: dot x=x-y+8t…
Solve the system of differential equations: $$ \dot x=x-y+8t,\qquad \dot y=5x-y $$
Explanation of Method
The homogeneous part is purely oscillatory (eigenvalues $\pm2i$), built from a complex eigenvector split into real and imaginary parts; since $\mu=0$ is not an eigenvalue, a linear-in-$t$ ansatz finds the particular solution directly.
- Write the system in matrix form and find the (purely imaginary) eigenvalues.
- Find the complex eigenvector and split it into real and imaginary parts.
- Assemble the homogeneous (oscillatory) solution.
- Find a particular solution of the form $\mathbf a\,t+\mathbf b$.
Step 1 · Characteristic equation
With $A=\begin{pmatrix}1&-1\\5&-1\end{pmatrix}$,
$$ \det(A-\lambda I)=\lambda^2+4=0 \implies \lambda=\pm2i. $$
Step 2 · Complex eigenvector and real solutions
For $\lambda=2i$: $\mathbf v=(1,1-2i)=\mathbf a+i\mathbf b$ with $\mathbf a=(1,1)$, $\mathbf b=(0,-2)$. The real solutions are $\mathbf a\cos2t-\mathbf b\sin2t$ and $\mathbf a\sin2t+\mathbf b\cos2t$.
Step 3 · Particular solution
Since $\mu=0$ is not an eigenvalue (as $\det A=4\neq0$), try $\mathbf u_p=\mathbf a t+\mathbf b$. Matching coefficients gives $\mathbf a=(2,10)$, $\mathbf b=(2,0)$:
$$ \mathbf u_p=\begin{pmatrix}2t+2\\10t\end{pmatrix}. $$
Step 4 · Verification
Substituting the general solution into both original equations confirms each holds identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda=\pm2i$ (purely oscillatory homogeneous part) |
| Forcing | $8t$ in the $x$-equation only |
| Particular solution | $(2t+2,\ 10t)$ |
Additional Notes
- Because the homogeneous solution is bounded (purely oscillatory), the long-term behaviour of the full solution is dominated entirely by the linearly growing particular solution.
- As with Problem 45, the purely imaginary eigenvalue pair means no resonance correction is needed for the polynomial forcing, since $\mu=0\neq\pm2i$.
- Verified by direct substitution into both original equations.
Result
The general solution of the system is: