IB Mathematics AHL 5.18 Solutions of second order differential equation AI HL Paper 2 - Exam Style Questions- New Syllabus

Question

A shock absorber on a car contains a spring surrounded by a fluid. When the car travels over uneven ground the spring is compressed and then returns to an equilibrium position.

The displacement, \( x \), of the spring is measured, in centimetres, from the equilibrium position of \( x = 0 \). The value of \( x \) can be modelled by the following second order differential equation, where \( t \) is the time, measured in seconds, after the initial displacement.
\[ \ddot{x} + 3\dot{x} + 1.25x = 0 \]

(a) Given that \( y = \dot{x} \), show that \( \dot{y} = -1.25x – 3y \).

(b) The differential equation can be expressed in the form \( \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix} \), where \( A \) is a \( 2 \times 2 \) matrix. Write down the matrix \( A \).

(c) (i) Find the eigenvalues of matrix \( A \).
(ii) Find the eigenvectors of matrix \( A \).

(d) Given that when \( t = 0 \) the shock absorber is displaced 8 cm and its velocity is zero, find an expression for \( x \) in terms of \( t \).

▶️ Answer/Explanation

Markscheme

(a)
    Given \( y = \dot{x} \Rightarrow \dot{y} = \ddot{x} \) (A1)
    From the differential equation \( \ddot{x} + 3\dot{x} + 1.25x = 0 \):
    \( \ddot{x} = -3\dot{x} – 1.25x \) (R1)
    Substitute \( \dot{x} = y \):
    \( \dot{y} = -1.25x – 3y \) (AG)
Result:
\( \dot{y} = -1.25x – 3y \)

(b)
    Express the system in matrix form \( \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix} \):
    \( \dot{x} = y \)
    \( \dot{y} = -1.25x – 3y \)
    So the matrix \( A = \begin{pmatrix} 0 & 1 \\ -1.25 & -3 \end{pmatrix} \) (A1)
Result:
\( A = \begin{pmatrix} 0 & 1 \\ -1.25 & -3 \end{pmatrix} \)

(c)(i)
    Form the characteristic equation:
    \( \det(A – \lambda I) = \begin{vmatrix} -\lambda & 1 \\ -1.25 & -3 – \lambda \end{vmatrix} = 0 \) (M1)
    Calculate the determinant:
    \( (-\lambda) \times (-3 – \lambda) – (1 \times -1.25) = 0 \)
    \( \lambda (3 + \lambda) + 1.25 = 0 \)
    \( \lambda^2 + 3\lambda + 1.25 = 0 \) (A1)
    Solve the quadratic equation:
    Discriminant = \( 3^2 – 4 \times 1 \times 1.25 = 9 – 5 = 4 \)
    \( \lambda = \frac{-3 \pm \sqrt{4}}{2} = \frac{-3 \pm 2}{2} \)
    \( \lambda = -2.5 \) or \( \lambda = -0.5 \) (A1)
Result:
Eigenvalues: \( \lambda = -2.5, -0.5 \)

(c)(ii)
    For \( \lambda = -2.5 \):
    \( (A + 2.5I)\mathbf{v} = 0 \Rightarrow \begin{pmatrix} 2.5 & 1 \\ -1.25 & -0.5 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \) (M1)
    \( 2.5v_1 + v_2 = 0 \Rightarrow v_2 = -2.5v_1 \)
    Choose \( v_1 = 1 \), so \( \mathbf{v} = \begin{pmatrix} 1 \\ -2.5 \end{pmatrix} \)
    Or scaled: \( \mathbf{v} = \begin{pmatrix} 2 \\ -5 \end{pmatrix} \) (A1)
    For \( \lambda = -0.5 \):
    \( (A + 0.5I)\mathbf{v} = 0 \Rightarrow \begin{pmatrix} 0.5 & 1 \\ -1.25 & -2.5 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \)
    \( 0.5v_1 + v_2 = 0 \Rightarrow v_2 = -0.5v_1 \)
    Choose \( v_1 = 1 \), so \( \mathbf{v} = \begin{pmatrix} 1 \\ -0.5 \end{pmatrix} \)
    Or scaled: \( \mathbf{v} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} \) (A1)
Result:
Eigenvectors: \( \begin{pmatrix} 2 \\ -5 \end{pmatrix} \) and \( \begin{pmatrix} 2 \\ -1 \end{pmatrix} \) (or equivalent multiples)

(d)
    Form the general solution:
    \( \begin{pmatrix} x \\ y \end{pmatrix} = C_1 e^{-2.5t} \begin{pmatrix} 2 \\ -5 \end{pmatrix} + C_2 e^{-0.5t} \begin{pmatrix} 2 \\ -1 \end{pmatrix} \) (M1)(A1)
    Apply initial conditions \( x(0) = 8 \), \( y(0) = 0 \) (since \( y = \dot{x} \) and velocity is zero):
    \( x(0) = 2C_1 + 2C_2 = 8 \)
    \( y(0) = -5C_1 – C_2 = 0 \) (M1)
    Solve the system:
    From second equation: \( -5C_1 – C_2 = 0 \Rightarrow C_2 = -5C_1 \)
    Substitute into first: \( 2C_1 + 2(-5C_1) = 8 \)
    \( 2C_1 – 10C_1 = 8 \Rightarrow -8C_1 = 8 \Rightarrow C_1 = -1 \)
    \( C_2 = -5 \times -1 = 5 \) (M1)(A1)
    So \( x(t) = -2e^{-2.5t} + 10e^{-0.5t} \) (A1)
Result:
\( x(t) = -2e^{-2.5t} + 10e^{-0.5t} \)

IB Mathematics AHL 5.18 Solutions of second order differential equation AI HL Paper 2 - Exam Style Questions- New Syllabus

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