Solve the system \[ \frac{dx}{dt}=-3x+10y, \qquad \frac{dy}{dt}=-3x+8y. \]
Solution
1. The system is\[ \frac{d}{dt} \begin{pmatrix} x\\ y \end{pmatrix} = A \begin{pmatrix} x\\ y \end{pmatrix}, \qquad A= \begin{pmatrix} -3&10\\ -3&8 \end{pmatrix}. \]
This is exactly the given pair of differential equations written in matrix form.
2. The eigenvalues of \(A\) are \(2\) and \(3\).
The characteristic polynomial is \[ \begin{aligned} \det(A-\lambda I) &= \det \begin{pmatrix} -3-\lambda&10\\ -3&8-\lambda \end{pmatrix}\\ &= (-3-\lambda)(8-\lambda)+30\\ &= \lambda^2-5\lambda+6\\ &= (\lambda-2)(\lambda-3). \end{aligned} \]
3. Eigenvectors for the eigenvalues \(2\) and \(3\) may be chosen as\[ v_2= \begin{pmatrix} 2\\ 1 \end{pmatrix}, \qquad v_3= \begin{pmatrix} 5\\ 3 \end{pmatrix}. \]
For \(\lambda=2\), \[ (A-2I) \begin{pmatrix} 2\\ 1 \end{pmatrix} = 0. \] For \(\lambda=3\), \[ (A-3I) \begin{pmatrix} 5\\ 3 \end{pmatrix} = 0. \]
4. The general solution is\[ \boxed{ \begin{pmatrix} x(t)\\ y(t) \end{pmatrix} = C_1e^{2t} \begin{pmatrix} 2\\ 1 \end{pmatrix} + C_2e^{3t} \begin{pmatrix} 5\\ 3 \end{pmatrix}, } \]where \(C_1,C_2\) are arbitrary constants.
For an eigenpair \((\lambda,v)\) of \(A\), the function \[ u(t)=e^{\lambda t}v \] satisfies \[ u'(t) = \lambda e^{\lambda t}v = e^{\lambda t}Av = Au(t). \] Thus the two displayed eigenmodes are solutions. Their eigenvectors are linearly independent because they correspond to distinct eigenvalues, so these two solutions form a basis of the two-dimensional solution space of the first-order system.
5. Equivalently,\[ \boxed{ x(t)=2C_1e^{2t}+5C_2e^{3t}, \qquad y(t)=C_1e^{2t}+3C_2e^{3t}. } \]
Read off the two coordinates in step 4.
6. Q.E.D.
Step 5 gives the complete family of solutions.