IB Mathematics AHL 5.14 Solving by separation of variables-AI HL Paper 2- Exam Style Questions- New Syllabus

(a)
To determine the eigenvalues, set up the characteristic equation by computing the determinant of \( \begin{pmatrix} -4 – \lambda & 6 \\ 9 & -1 – \lambda \end{pmatrix} \) and equating it to zero:
\( \begin{vmatrix} -4 – \lambda & 6 \\ 9 & -1 – \lambda \end{vmatrix} = 0 \)
Expand the determinant:
\( (-4 – \lambda)(-1 – \lambda) – (6 \times 9) = 0 \)
\( (-4 – \lambda)(-1 – \lambda) – 54 = 0 \)
\( (4 + \lambda)(1 + \lambda) – 54 = 0 \)
\( 4 + 4\lambda + \lambda + \lambda^2 – 54 = 0 \)
\( \lambda^2 + 5\lambda – 50 = 0 \)
Solve using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = 1 \), \( b = 5 \), \( c = -50 \):
\( \lambda = \frac{-5 \pm \sqrt{5^2 – 4 \cdot 1 \cdot (-50)}}{2} = \frac{-5 \pm \sqrt{25 + 200}}{2} = \frac{-5 \pm \sqrt{225}}{2} = \frac{-5 \pm 15}{2} \)
Thus, the eigenvalues are:
\( \lambda_1 = \frac{-5 – 15}{2} = -10 \), \( \lambda_2 = \frac{-5 + 15}{2} = 5 \)
(Since \( \lambda_2 > \lambda_1 \), \( \lambda_1 = -10 \), \( \lambda_2 = 5 \).)
For \( \lambda_1 = -10 \):
Solve \( \begin{pmatrix} -4 – (-10) & 6 \\ 9 & -1 – (-10) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \), i.e.,
\( \begin{pmatrix} 6 & 6 \\ 9 & 9 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \)
From the first row, \( 6x + 6y = 0 \implies x + y = 0 \implies x = -y \).
A possible eigenvector \( p_1 \) is \( \begin{pmatrix} -1 \\ 1 \end{pmatrix} \).
For \( \lambda_2 = 5 \):
Solve \( \begin{pmatrix} -4 – 5 & 6 \\ 9 & -1 – 5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \), i.e.,
\( \begin{pmatrix} -9 & 6 \\ 9 & -6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \)
From the first row, \( -9x + 6y = 0 \implies 3x = 2y \implies x = \frac{2}{3}y \).
A possible eigenvector \( p_2 \) is \( \begin{pmatrix} 2 \\ 3 \end{pmatrix} \) (scaled for simplicity).

(b)
Using the eigenvalues and eigenvectors, the general solution of the system, in the required form, is:
\( \begin{pmatrix} x \\ y \end{pmatrix} = A p_1 e^{\lambda_1 t} + B p_2 e^{\lambda_2 t} \)
Substituting \( \lambda_1 = -10 \), \( p_1 = \begin{pmatrix} -1 \\ 1 \end{pmatrix} \), \( \lambda_2 = 5 \), \( p_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \):
\( \begin{pmatrix} x \\ y \end{pmatrix} = A e^{-10t} \begin{pmatrix} -1 \\ 1 \end{pmatrix} + B e^{5t} \begin{pmatrix} 2 \\ 3 \end{pmatrix} \)
where \( A \) and \( B \) are scalar constants.

(c)
At \( t = 0 \), the initial conditions are \( x = 500 \) and \( y = 125 \).
Substitute into the general solution:
\( x = -A + 2B = 500 \)
\( y = A + 3B = 125 \)
Solve the system of equations:
From the second equation, \( A = 125 – 3B \).
Substitute into the first equation:
\( -(125 – 3B) + 2B = 500 \)
\( -125 + 3B + 2B = 500 \)
\( 5B – 125 = 500 \)
\( 5B = 625 \implies B = 125 \)
Then, \( A = 125 – 3 \times 125 = 125 – 375 = -250 \).
Thus, \( A = -250 \) and \( B = 125 \).
The specific solution is:
\( \begin{pmatrix} x \\ y \end{pmatrix} = -250 e^{-10t} \begin{pmatrix} -1 \\ 1 \end{pmatrix} + 125 e^{5t} \begin{pmatrix} 2 \\ 3 \end{pmatrix} \)

(d)
The long-term behavior is dominated by the positive eigenvalue \( \lambda_2 = 5 \), corresponding to the eigenvector \( p_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \).
As \( t \to \infty \), the term with \( e^{-10t} \) decays to zero, and the solution approaches the direction of \( \begin{pmatrix} 2 \\ 3 \end{pmatrix} \).
The ratio of the components \( x \) to \( y \) is \( 2:3 \).
Thus, the long-term ratio of male to female fruit flies is \( 2:3 \).

(e)
To find \( \frac{dy}{dx} \) at \( t = 0 \), use the relationship derived from the differential equations:
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{9x – y}{-4x + 6y} \)
At \( t = 0 \), \( x = 500 \), \( y = 125 \):
\( \frac{dy}{dx} = \frac{9 \times 500 – 125}{-4 \times 500 + 6 \times 125} = \frac{4500 – 125}{-2000 + 750} = \frac{4375}{-1250} \)
\( \frac{dy}{dx} = \frac{4375}{-1250} = -3.5 \)
The value of \( \frac{dy}{dx} \) at \( t = 0 \) is \( -3.5 \).

(f)
The trajectory on the phase portrait reflects the system’s dynamics:
– Start at the initial point \( (500, 125) \) with an initial slope of \( -3.5 \), as calculated.
– The negative eigenvalue \( \lambda_1 = -10 \) (eigenvector \( \begin{pmatrix} -1 \\ 1 \end{pmatrix} \)) indicates decay, while the positive eigenvalue \( \lambda_2 = 5 \) (eigenvector \( \begin{pmatrix} 2 \\ 3 \end{pmatrix} \)) drives growth.
– The trajectory tends toward an oblique asymptote along the direction of \( \begin{pmatrix} 2 \\ 3 \end{pmatrix} \), which corresponds to the line \( y = \frac{3}{2}x \).
– Sketch a curve starting at \( (500, 125) \), initially sloping downward (due to \( -3.5 \)), and asymptotically approaching \( y = \frac{3}{2}x \) as \( t \) increases.