Question

An ordinary differential equation (ODE), \( \dfrac{dy}{dx} = 2y\), with an initial condition \(y(0) = 1\), has the analytical solution \(y = e^{2x}\). Using Runge-Kutta second order method, numerically integrate the ODE to calculate y at \(x = 0.5\) using a step size of \(h = 0.5\). 
If the relative percentage error is defined as, \[ \varepsilon = \left| \frac{y_{\text{analytical}} - y_{\text{numerical}}}{y_{\text{analytical}}} \right| \times 100, \] then the value of \(\varepsilon\) at \(x = 0.5\) is \(\underline{\hspace{2cm}}\). 
 

• A. 0.06
• B. 0.8
• C. 4.0
• D. 8.0

Hint:

In RK-2 midpoint method, always compute $k_2$ at the half-step. For exponential growth ODEs, RK-2 tends to underestimate, leading to noticeable error for large step sizes.

Answer 1

The Correct Option is D

We are given the ODE: \[ \frac{dy}{dx} = 2y, y(0)=1. \]

Step 1: Analytical solution.
\[ y(x) = e^{2x}. \] At \(x = 0.5\): \[ y_{\text{analytical}} = e^{1} = 2.71828. \]

Step 2: Apply RK-2 method (midpoint form).
The RK-2 method uses: \[ k_1 = f(x_0,y_0) \] \[ k_2 = f(x_0 + h/2, y_0 + h k_1/2) \] \[ y_{1} = y_0 + h\, k_2 \] Given: \[ f(x,y)=2y, y_0 = 1, h = 0.5. \] Compute \(k_1\): \[ k_1 = 2y_0 = 2(1)=2. \] Compute \(k_2\): \[ y_0 + \frac{h k_1}{2} = 1 + \frac{0.5 \times 2}{2} = 1 + 0.5 = 1.5. \] \[ k_2 = 2(1.5)=3. \] Now compute numerical value: \[ y_{1} = y_0 + h k_2 = 1 + 0.5 \times 3 = 2.5. \]

Step 3: Compute percentage error.
\[ \varepsilon = \left| \frac{y_{\text{analytical}} - y_{\text{numerical}}}{y_{\text{analytical}}} \right| \times 100 \] \[ \varepsilon = \left| \frac{2.71828 - 2.5}{2.71828} \right| \times 100 \] \[ = \left| \frac{0.21828}{2.71828} \right| \times 100 = 0.08025 \times 100 \approx 8.0 \]

Step 4: Conclusion.
The relative percentage error is approximately \(8%\).