Q2Advanced Engineering Mathematics

Question

If \frac{d y}{d x} = x + y^{2} use Runge-Kutta method to find an approximate value of y for x = 0.2, given that y = 1 when x = 0. Use Laplace transform to solve.

Answer 1

The Runge-Kutta family of algorithms represents the gold standard for numerically solving Ordinary Differential Equations (ODEs) within engineering applications. While single-step Euler methods are conceptually simple, their high truncation error mandates impractically minuscule step sizes. Multi-step methods like Milne's are highly accurate but suffer from the severe disadvantage of not being self-starting. The 4th Order Runge-Kutta method (RK4) beautifully balances these constraints: it is a robust, self-starting single-step method that achieves exceptional fourth-order accuracy ( $O (h^{4})$ ) by strategically probing the derivative function at multiple intermediate points within each interval.

1. Theoretical Architecture of RK4

Consider the standard Initial Value Problem (IVP): $\frac{d y}{d x} = f (x, y)$ subject to the strict initial condition $y (x_{0}) = y_{0}$ . We intend to step forward by a finite increment $h$ to find the corresponding value $y_{1}$ at $x_{1} = x_{0} + h$ .

The RK4 algorithm constructs a weighted average of four distinct, carefully sampled slopes ( $k_{1}, k_{2}, k_{3}, k_{4}$ ) across the interval to predict the final trajectory.

$k_{1}$ (Initial Slope): This is the exact slope evaluated exactly at the starting node $(x_{0}, y_{0})$ . It is essentially a standard Euler prediction. $k_{1} = h \cdot f (x_{0}, y_{0})$ .
$k_{2}$ (First Midpoint Slope): The algorithm uses $k_{1}$ to project halfway across the interval (to $x_{0} + h /2$ ) and calculates the slope at this tentatively estimated midpoint. $k_{2} = h \cdot f (x_{0} + \frac{h}{2}, y_{0} + \frac{k _{1}}{2})$ .
$k_{3}$ (Second Midpoint Slope): The algorithm refines its estimate of the midpoint by utilizing the newly calculated $k_{2}$ slope to project from the start. It recalculates the slope there. $k_{3} = h \cdot f (x_{0} + \frac{h}{2}, y_{0} + \frac{k _{2}}{2})$ .
$k_{4}$ (Terminal Slope): Finally, the algorithm utilizes $k_{3}$ to project all the way to the end of the interval (to $x_{0} + h$ ) and evaluates the slope at this final predicted boundary. $k_{4} = h \cdot f (x_{0} + h, y_{0} + k_{3})$ .

The final formula ingeniously combines these four slopes using a weighted average derived precisely from Simpson's 1/3 integration rule, heavily favoring the mathematically superior midpoint estimations (weighting them twice as heavily as the boundary slopes):

y_{1} = y_{0} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})

2. Step-by-Step Execution for the Given Problem

The specific mathematical problem presented is $\frac{d y}{d x} = x + y^{2}$ , with initial boundaries $x_{0} = 0$ , $y_{0} = 1$ , and a specified step size $h = 0.2$ . The objective function is $f (x, y) = x + y^{2}$ .

Calculating $k_{1}$ :

k_{1} = (0.2) \cdot [(0) + (1)^{2}] = (0.2) \cdot [1] = 0.2000

Calculating $k_{2}$ : The coordinates shift to $x = 0 + 0.1 = 0.1$ and $y = 1 + 0.2/2 = 1.1$ .

k_{2} = (0.2) \cdot [(0.1) + (1.1)^{2}] = (0.2) \cdot [0.1 + 1.21] = (0.2) \cdot [1.31] = 0.2620

Calculating $k_{3}$ : The coordinates shift to $x = 0 + 0.1 = 0.1$ and $y = 1 + 0.262/2 = 1.131$ .

k_{3} = (0.2) \cdot [(0.1) + (1.131)^{2}] = (0.2) \cdot [0.1 + 1.279161] = (0.2) \cdot [1.379161] = 0.2758322

Calculating $k_{4}$ : The coordinates shift to the terminal boundary $x = 0 + 0.2 = 0.2$ and $y = 1 + 0.2758322 = 1.2758322$ .

k_{4} = (0.2) \cdot [(0.2) + (1.2758322)^{2}] = (0.2) \cdot [0.2 + 1.627747] = (0.2) \cdot [1.827747] = 0.3655494

3. Final Synthesis

With all four constituent slope approximations meticulously calculated to high precision, we blend them into the final Simpson's weighting equation to determine the ultimate function value at the new coordinate $x = 0.2$ .

y (0.2) = y_{0} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})

y (0.2) = 1.0 + \frac{1}{6} (0.2000 + 2 (0.2620) + 2 (0.2758322) + 0.3655494)

y (0.2) = 1.0 + \frac{1}{6} (0.2000 + 0.5240 + 0.5516644 + 0.3655494) = 1.0 + \frac{1}{6} (1.6412138)

y (0.2) = 1.0 + 0.2735356 \approx 1.2735

Thus, the highly accurate approximated solution for the non-linear differential equation evaluated precisely at $x = 0.2$ is $1.2735$ .

Q2Advanced Engineering Mathematics

Question

Answer

1. Theoretical Architecture of RK4

2. Step-by-Step Execution for the Given Problem

3. Final Synthesis