Created on 2024-06-20Asked by Emma Lewis (Solvelet student)
Solve the initial value problem y′=x+y, y(0)=1 using the Runge-Kutta method of order 4 with step size h=0.1 to approximate y(0.1).
Solution
To solve the initial value problem y′=x+y, y(0)=1 using the Runge-Kutta method of order 4 with step size h=0.1: 1. **Define the function:** f(x,y)=x+y. 2. **Calculate the Runge-Kutta coefficients:** \begin{align*} k_1 &= h f(x_n, y_n) = 0.1 (0 + 1) = 0.1, \\ k_2 &= h f\left(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}\right) = 0.1 \left(0.05 + 1.05\right) = 0.115, \\ k_3 &= h f\left(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}\right) = 0.1 \left(0.05 + 1.0575\right) = 0.11075, \\ k_4 &= h f(x_n + h, y_n + k_3) = 0.1 \left(0.1 + 1.11075\right) = 0.121075. \end{align*} 3. **Update the value of y:** yn+1=yn+6k1+2k2+2k3+k4=1+60.1+2⋅0.115+2⋅0.11075+0.121075=1.110625. 4. **Result:** The approximate value of y(0.1) using the Runge-Kutta method of order 4 is 1.110625. Solved on Solvelet with Basic AI Model
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DefinitionRunge-Kutta methods are iterative possible ways of solving ordinary differential equations. For the fourth-order method, it is the most common because it has this trade-off between accuracy and computationally expensive. For: Runge-Kutta 4th Order Method k1 = Δty´(t)k2 = Δty´(t + Δt/2)k3 = Δty´(t + Δt/2)k4 = Δty´(t + Δt)y(t + Δt) = y(t) + 1/6(k1 + 2k2 + 2k3 + k4) — k1rpm
