Question

The solution \(y(x) = \binom{y_1(x)}{y_2(x)}\) of the initial value problem \[ \frac{dy}{dx} = \begin{pmatrix} -3 & 4 \\ -2 & 3 \end{pmatrix} y, \qquad y(0) = \binom{1}{2} \] is equal to

• A. \(3\dbinom{1}{1}e^{x} + \dbinom{2}{1}e^{-x}\)
• B. \(-3\dbinom{1}{1}e^{x} + \dbinom{2}{1}e^{-x}\)
• C. \(3\dbinom{1}{1}e^{x} + 2\dbinom{2}{1}e^{-x}\)
• D. \(3\dbinom{1}{1}e^{x} - \dbinom{2}{1}e^{-x}\)

Hint:

Find the eigenvalues and eigenvectors of the coefficient matrix, then fix the constants using the given initial vector \(y(0)=\binom{1}{2}\).

Answer 1

The Correct Option is D

A different way to solve this system is to eliminate one variable and get a single second order equation for $y_1$, instead of working with eigenvectors directly.

The system is $y_1' = -3y_1+4y_2$ and $y_2' = -2y_1+3y_2$, with $y_1(0)=1$, $y_2(0)=2$.

From the first equation, $y_2 = \dfrac{y_1'+3y_1}{4}$. Differentiating this gives $y_2' = \dfrac{y_1''+3y_1'}{4}$.

Substitute both expressions into the second equation:

\[ \frac{y_1''+3y_1'}{4} = -2y_1 + 3\left(\frac{y_1'+3y_1}{4}\right) \]

Multiply through by 4: $y_1''+3y_1' = -8y_1+3y_1'+9y_1 = 3y_1'+y_1$. The $3y_1'$ terms cancel, leaving $y_1''=y_1$, so $y_1''-y_1=0$.

The characteristic roots are $r=\pm 1$, so $y_1(x) = Ae^{x}+Be^{-x}$ for constants $A,B$.

Using $y_1(0)=1$ gives $A+B=1$. Differentiating, $y_1'(x)=Ae^{x}-Be^{-x}$, and from the first equation $y_1'(0) = -3y_1(0)+4y_2(0) = -3(1)+4(2) = 5$, so $A-B=5$.

Solving these two equations, $2A=6$, so $A=3$ and $B=-2$. This gives $y_1(x) = 3e^{x}-2e^{-x}$.

Now find $y_2$ using $y_2 = \dfrac{y_1'+3y_1}{4}$. Here $y_1' = 3e^{x}+2e^{-x}$, so $y_1'+3y_1 = 3e^{x}+2e^{-x}+9e^{x}-6e^{-x} = 12e^{x}-4e^{-x}$, and dividing by 4 gives $y_2(x) = 3e^{x}-e^{-x}$.

Writing $y_1$ and $y_2$ together as a vector, $y(x) = 3\binom{1}{1}e^{x} - \binom{2}{1}e^{-x}$, which matches the same coefficients found by the eigenvector method, confirming option (D) is correct.

\[ \boxed{y(x)=3\binom{1}{1}e^{x}-\binom{2}{1}e^{-x}} \]