Solve the system \[ \frac{dy_1}{dx}=-3y_1+10y_2, \qquad \frac{dy_2}{dx}=-3y_1+8y_2. \]
Solution
Write \[ Y= \begin{pmatrix} y_1\\ y_2 \end{pmatrix}, \qquad M= \begin{pmatrix} -3&10\\ -3&8 \end{pmatrix}. \] Then the system is \(Y'=MY\).
1. The eigenvalues of \(M\) are \(2\) and \(3\).
The characteristic polynomial is \[ \begin{aligned} \det(\lambda I-M) &= \det \begin{pmatrix} \lambda+3&-10\\ 3&\lambda-8 \end{pmatrix}\\ &= (\lambda+3)(\lambda-8)+30\\ &= \lambda^2-5\lambda+6\\ &= (\lambda-2)(\lambda-3). \end{aligned} \]
2. Eigenvectors for the eigenvalues \(2\) and \(3\) are respectively\[ v_2= \begin{pmatrix} 2\\ 1 \end{pmatrix}, \qquad v_3= \begin{pmatrix} 5\\ 3 \end{pmatrix}. \]
Direct multiplication gives \[ Mv_2 = \begin{pmatrix} 4\\ 2 \end{pmatrix} = 2v_2 \] and \[ Mv_3 = \begin{pmatrix} 15\\ 9 \end{pmatrix} = 3v_3. \]
3. The vectors \(v_2,v_3\) form a basis of \(\RR^2\).
Their determinant is \[ \det \begin{pmatrix} 2&5\\ 1&3 \end{pmatrix} = 1\neq0. \] Hence they are linearly independent and therefore form a basis.
4. If\[ Y(x)=c_1(x)v_2+c_2(x)v_3, \]then the system \(Y'=MY\) is equivalent to\[ c_1'=2c_1, \qquad c_2'=3c_2. \]
Differentiating the basis expansion gives \[ Y' = c_1'v_2+c_2'v_3. \] By step 2, \[ MY = c_1Mv_2+c_2Mv_3 = 2c_1v_2+3c_2v_3. \] Since \(v_2,v_3\) are a basis by step 3, equality \(Y'=MY\) holds exactly when the corresponding coefficients agree.
5. The general solution is\[ \boxed{ Y(x) = C_1e^{2x} \begin{pmatrix} 2\\ 1 \end{pmatrix} + C_2e^{3x} \begin{pmatrix} 5\\ 3 \end{pmatrix} }, \qquad C_1,C_2\in\RR. \]
The scalar equations in step 4 have the general solutions \[ c_1(x)=C_1e^{2x}, \qquad c_2(x)=C_2e^{3x}. \] Substituting these into the basis expansion gives the displayed formula. Conversely, every displayed function satisfies the two scalar equations and hence, by step 4, the original system.
6. Equivalently,\[ \boxed{ \begin{aligned} y_1(x)&=2C_1e^{2x}+5C_2e^{3x},\\ y_2(x)&=C_1e^{2x}+3C_2e^{3x}. \end{aligned} } \]
This is the coordinate form of the vector solution in step 5.
7. Q.E.D.
Steps 5 and 6 give all solutions of the system.