Solve the system of differential equations: (dy)/(dx) = y +…
Solve the system of differential equations: $$ \frac{dy}{dx} = y + z + x,\qquad \frac{dz}{dx} = -4y - 3z + 2x $$
Explanation of Method
Write the system as $\mathbf{u}'=A\mathbf{u}+\mathbf{f}(x)$. Since $A$ has a repeated eigenvalue, the homogeneous solution needs a generalised eigenvector; the linear forcing term is matched with a linear-in-$x$ particular solution.
- Form the matrix $A$ and solve the characteristic equation.
- Find the eigenvector, and a generalised eigenvector if the eigenvalue repeats.
- Assemble the homogeneous solution.
- Guess a particular solution matching the forcing term and solve for its coefficients.
- Add homogeneous and particular parts for the general solution.
Step 1 · Matrix form and characteristic equation
With $A=\begin{pmatrix}1&1\\-4&-3\end{pmatrix}$, $\mathbf{f}(x)=\begin{pmatrix}x\\2x\end{pmatrix}$, the characteristic equation is
$$ \det(A-\lambda I)=(1-\lambda)(-3-\lambda)+4=(\lambda+1)^2=0, $$
so $\lambda=-1$ is a repeated eigenvalue.
Step 2 · Eigenvector and generalised eigenvector
Solving $(A+I)\mathbf v=0$ gives the eigenvector, and $(A+I)\mathbf v_2=\mathbf v_1$ gives the generalised eigenvector:
$$ \mathbf v_1=\begin{pmatrix}1\\-2\end{pmatrix},\qquad \mathbf v_2=\begin{pmatrix}0\\1\end{pmatrix}. $$
Step 3 · Homogeneous solution
$$ \begin{pmatrix}y_h\\z_h\end{pmatrix}=C_1e^{-x}\begin{pmatrix}1\\-2\end{pmatrix}+C_2e^{-x}\begin{pmatrix}x\\-2x+1\end{pmatrix}. $$
Step 4 · Particular solution
Trying $\mathbf u_p=(ax+b,\,cx+d)^{\mathsf T}$ and matching coefficients of $x$ and constants gives
$$ \mathbf u_p=\begin{pmatrix}5x-9\\-6x+14\end{pmatrix}. $$
Step 5 · General solution
$$ y=C_1e^{-x}+C_2xe^{-x}+5x-9,\qquad z=-2C_1e^{-x}+C_2(1-2x)e^{-x}-6x+14. $$
Step 6 · Verification
Differentiating $y$ and $z$ and substituting back into both equations confirms $y'=y+z+x$ and $z'=-4y-3z+2x$ identically. ✓
Summary Table
| Quantity | Value |
|---|---|
| Eigenvalue | $\lambda=-1$ (multiplicity 2) |
| Eigenvector | $(1,-2)$ |
| Generalised eigenvector | $(0,1)$ |
| Particular solution | $(5x-9,\,-6x+14)$ |
Additional Notes
- A repeated eigenvalue with a one-dimensional eigenspace always requires a generalised (chain) eigenvector to build a second independent solution.
- Because $\mathbf f(x)$ is linear and $0$ is not an eigenvalue of $A$, a linear ansatz for the particular solution is guaranteed to work.
- The result was checked by direct substitution into both original equations.
Result
The general solution of the system is: