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A) Discuss the i) theory of classical condition (10 marks) and ii) the theory of operant condition (10 marks). b) Discuss how the two theories can be applied in a life of learner (20 marks).

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Answer

b

x(t)=x0\cn(Ω(t−t0)∣k)x(t) = x_0 \cn(\Omega(t - t_0) \mid k)

Step-by-step solution

9 steps

Part a· 5 steps

Classical (Exact) Solution (20 marks)

  1. Identify the differential equation.

    The Duffing equation (undamped, unforced case) is

    x¨+αx+βx3=0,\ddot{x} + \alpha x + \beta x^3 = 0,

    where typically α>0\alpha > 0 (hardening or softening spring depending on β\beta).

  2. Multiply by x˙\dot{x} to get energy conservation.

    x˙x¨+αxx˙+βx3x˙=0.\dot{x} \ddot{x} + \alpha x \dot{x} + \beta x^3 \dot{x} = 0.

    Integrate with respect to tt:

    12x˙2+α2x2+β4x4=E,\frac{1}{2} \dot{x}^2 + \frac{\alpha}{2} x^2 + \frac{\beta}{4} x^4 = E,

    where EE is a constant (total energy).

  3. Separate variables.

    From the energy equation,

    x˙=±2(E−α2x2−β4x4).\dot{x} = \pm \sqrt{2 \left( E - \frac{\alpha}{2} x^2 - \frac{\beta}{4} x^4 \right)}.

    Thus,

    dt=dx±2(E−α2x2−β4x4).dt = \frac{dx}{\pm \sqrt{2 \left( E - \frac{\alpha}{2} x^2 - \frac{\beta}{4} x^4 \right)}}.
  4. Integrate to find the period or solution form.

    The solution is periodic. The quarter-period T/4T/4 is

    T4=∫0Adx2(E−α2x2−β4x4),\frac{T}{4} = \int_0^{A} \frac{dx}{\sqrt{2 \left( E - \frac{\alpha}{2} x^2 - \frac{\beta}{4} x^4 \right)}},

    where AA is the amplitude (E=α2A2+β4A4E = \frac{\alpha}{2} A^2 + \frac{\beta}{4} A^4).

  5. Recognize elliptic integral form.

    This is an elliptic integral of the first kind. The exact solution is

    x(t)=x0 cn(Ω(t−t0)∣k),x(t) = x_0 \, \mathrm{cn} \left( \Omega (t - t_0) \mid k \right),

    where cn\mathrm{cn} is the Jacobi elliptic cosine function, and parameters Ω\Omega, kk depend on α\alpha, β\beta, x0x_0 (amplitude):

    k=βx022(α+βx02),Ω=α+βx021+k2.k = \sqrt{\frac{\beta x_0^2}{2(\alpha + \beta x_0^2)}}, \quad \Omega = \sqrt{\alpha + \beta x_0^2} \sqrt{1 + k^2}.

    (This is the classical exact solution using elliptic functions.)

Part b

x(t)=x0\cn(Ω(t−t0)∣k)x(t) = x_0 \cn(\Omega(t - t_0) \mid k)

Any Two Classical Methods (10 marks)

Method 1: Euler's Method

  1. Rewrite as system

    Let v=x˙v = \dot{x}, then

    x˙=v,v˙=−αx−βx3.\dot{x} = v, \quad \dot{v} = -\alpha x - \beta x^3.
  2. Euler update

    xn+1=xn+hvn,x_{n+1} = x_n + h v_n, vn+1=vn+h(−αxn−βxn3),v_{n+1} = v_n + h (-\alpha x_n - \beta x_n^3),

    where hh is step size.
    Simple, explicit, first-order accurate, but unstable for stiff problems.

    Method 2: Classical Fourth-Order Runge-Kutta (RK4)

  3. Same system y′=f(t,y)\mathbf{y}' = \mathbf{f}(t, \mathbf{y}), y=[x,v]T\mathbf{y} = [x, v]^T.

  4. RK4 increments

    k1=hf(tn,yn),k_1 = h \mathbf{f}(t_n, \mathbf{y}_n), k2=hf(tn+h2,yn+k12),k_2 = h \mathbf{f}(t_n + \frac{h}{2}, \mathbf{y}_n + \frac{k_1}{2}), k3=hf(tn+h2,yn+k22),k_3 = h \mathbf{f}(t_n + \frac{h}{2}, \mathbf{y}_n + \frac{k_2}{2}), k4=hf(tn+h,yn+k3).k_4 = h \mathbf{f}(t_n + h, \mathbf{y}_n + k_3).

    Then

    yn+1=yn+16(k1+2k2+2k3+k4).\mathbf{y}_{n+1} = \mathbf{y}_n + \frac{1}{6} (k_1 + 2k_2 + 2k_3 + k_4).

    Fourth-order accurate, good for nonlinear ODEs like Duffing.

    x(t) = x_0 \cn(\Omega(t - t_0) \mid k)

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