Solve the initial value problem: dot x=x+y+t, dot y=x-2y+2t…
Solve the initial value problem: $$ \dot x=x+y+t,\qquad \dot y=x-2y+2t,\qquad x(0)=-\tfrac79,\ y(0)=-\tfrac59 $$
Explanation of Method
The coefficient matrix here has irrational eigenvalues $\lambda=\tfrac{-1\pm\sqrt{13}}{2}$, which would normally make the general solution unwieldy; remarkably, the specific initial data given exactly matches the particular (polynomial) solution, so both homogeneous constants vanish and the messy exponential terms never appear in the final answer.
- Find the (irrational) eigenvalues of $A$ and the polynomial particular solution.
- Assemble the general solution with the irrational-exponent homogeneous terms.
- Apply the initial conditions and observe that both homogeneous constants come out to zero.
- State and verify the resulting purely polynomial particular solution.
Step 1 · Eigenvalues and particular solution
With $A=\begin{pmatrix}1&1\\1&-2\end{pmatrix}$, $\det(A-\lambda I)=\lambda^2+\lambda-3=0$ gives $\lambda_{1,2}=\dfrac{-1\pm\sqrt{13}}{2}$. Trying $\mathbf u_p=\mathbf p+\mathbf q\,t$ for the linear forcing gives
$$ \mathbf q=\begin{pmatrix}-\tfrac43\\[2pt]\tfrac13\end{pmatrix},\qquad \mathbf p=\begin{pmatrix}-\tfrac79\\[2pt]-\tfrac59\end{pmatrix}. $$
Step 2 · Apply the initial conditions
Since $x(0)=-\tfrac79$ and $y(0)=-\tfrac59$ exactly match $\mathbf p$, both initial-condition equations reduce to a homogeneous $2\times2$ system for $C_1,C_2$ whose only solution is
$$ C_1=C_2=0. $$
Step 3 · Particular solution
With both homogeneous constants zero, the irrational eigenvalues never appear, and the exact solution is simply
$$ x(t)=-\tfrac43t-\tfrac79,\qquad y(t)=\tfrac13t-\tfrac59. $$
Step 4 · Verification
Both initial conditions and both original differential equations are satisfied identically by direct substitution. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalues | $\lambda_{1,2}=\dfrac{-1\pm\sqrt{13}}{2}$ (irrational, but ultimately unused) |
| Constants from ICs | $C_1=C_2=0$ |
| Final solution | purely linear in $t$ |
Additional Notes
- This is a special (non-generic) case: the given initial data lies exactly on the particular solution, so the irrational-eigenvalue homogeneous modes are never excited, leaving a clean linear answer.
- For any other initial data, the general solution would genuinely involve $e^{\lambda_1t}$ and $e^{\lambda_2t}$ with irrational exponents.
- Verified against both given initial values and both original differential equations.
Result
The particular solution satisfying the initial data is: