--- title: "Program for RK-4 second order IVP" subject: "Numerical Method" chapter: 6 key: "semester_4/numerical_method:rk4-program-second-order" source: https://www.bctengineer.com/q/semester_4/numerical_method/rk4-program-second-order site: BCT Engineering --- # Program for RK-4 second order IVP Numerical Method · Chapter 6: Solution of ordinary differential equations - Asked in: 2073 Magh, 6 marks - Video: [RK4 2nd order ODE | Numerical Methods | LetThereBeMath |](https://www.youtube.com/watch?v=TjZgQa2kec0) — Let there be math - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/rk4-program-second-order - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-6 ## Question Write a program in C/C++/FORTRAN to solve a second order ordinary differential equation (initial value problem) using the Runge-Kutta fourth order method. ## Short answer The code reduces y''= g(x,y,y') to first-order system: dy/dx=z, dz/dx=g(x,y,z). RK4 is applied simultaneously: compute k1..k4 for y and l1..l4 for z using half-step evaluations, then update y and z. It prints x, y, y' each step. ## Answer A second-order IVP y'' = g(x, y, y') with y(x₀) = y₀, y'(x₀) = z₀ is reduced to two first-order equations by putting z = y': ```text dy/dx = z = f₁(x, y, z) dz/dx = g(x, y, z) = f₂(x, y, z) ``` RK-4 is then applied to both equations together: ```text k₁ = h·f₁(x, y, z) l₁ = h·f₂(x, y, z) k₂ = h·f₁(x+h/2, y+k₁/2, z+l₁/2) l₂ = h·f₂(x+h/2, y+k₁/2, z+l₁/2) k₃ = h·f₁(x+h/2, y+k₂/2, z+l₂/2) l₃ = h·f₂(x+h/2, y+k₂/2, z+l₂/2) k₄ = h·f₁(x+h, y+k₃, z+l₃) l₄ = h·f₂(x+h, y+k₃, z+l₃) y ← y + (k₁+2k₂+2k₃+k₄)/6 z ← z + (l₁+2l₂+2l₃+l₄)/6 ``` #### C program (example: y'' = y − x·y') ```c #include #include /* y'' = g(x, y, z) with z = y'. Example: y'' = y - x*y' */ double f1(double x, double y, double z) /* dy/dx = z */ { return z; } double f2(double x, double y, double z) /* dz/dx = y'' */ { return y - x * z; } int main(void) { double x0, y0, z0, xn, h, x, y, z; double k1, k2, k3, k4, l1, l2, l3, l4; int i, n; printf("Enter x0, y(x0), y'(x0), xn and h: "); scanf("%lf %lf %lf %lf %lf", &x0, &y0, &z0, &xn, &h); n = (int)((xn - x0) / h + 0.5); x = x0; y = y0; z = z0; printf("\n x y y'\n"); printf("%8.4f %10.4f %10.4f\n", x, y, z); for (i = 0; i < n; i++) { k1 = h * f1(x, y, z); l1 = h * f2(x, y, z); k2 = h * f1(x + h / 2, y + k1 / 2, z + l1 / 2); l2 = h * f2(x + h / 2, y + k1 / 2, z + l1 / 2); k3 = h * f1(x + h / 2, y + k2 / 2, z + l2 / 2); l3 = h * f2(x + h / 2, y + k2 / 2, z + l2 / 2); k4 = h * f1(x + h, y + k3, z + l3); l4 = h * f2(x + h, y + k3, z + l3); y = y + (k1 + 2 * k2 + 2 * k3 + k4) / 6; z = z + (l1 + 2 * l2 + 2 * l3 + l4) / 6; x = x + h; printf("%8.4f %10.4f %10.4f\n", x, y, z); } return 0; } ``` #### Sample run Input `0 1 0 0.2 0.1` (y(0) = 1, y'(0) = 0, up to x = 0.2, h = 0.1): ```text x y y' 0.0000 1.0000 0.0000 0.1000 1.0050 0.0998 0.2000 1.0199 0.1987 ``` For any other equation change only the `f2` function (the right side y'' = …). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a second-order IVP y'' = g(x, y, y') converted for RK-4? A. Apply RK-4 directly to y'' treating it as dy/dx B. Integrate y'' twice analytically before coding C. Put z = y', giving dy/dx = z and dz/dx = g(x, y, z), and apply RK-4 to both together D. Replace y'' by a forward difference and apply Euler's method Answer: C. Setting z = y' turns the equation into two first-order equations, which RK-4 can solve simultaneously. (Key point: Reduction to a system) 2. In the program, how are y and z updated at the end of each step? A. y += (k1+k2+k3+k4)/4 and z += (l1+l2+l3+l4)/4 B. y += k4 and z += l4 C. y += (l1+2l2+2l3+l4)/6 and z += (k1+2k2+2k3+k4)/6 D. y += (k1+2k2+2k3+k4)/6 and z += (l1+2l2+2l3+l4)/6 Answer: D. Each variable uses its own weighted average (1,2,2,1)/6 of its slopes: k for y and l for z. (Key point: Updating y and z) 3. To solve a different second-order equation with the same program, what must be changed? A. Only the function f2 that gives the right side y'' = ... B. Only the function f1 C. Both f1 and the RK-4 loop formulas D. The step size formula n = (xn-x0)/h Answer: A. f1 = z is always dy/dx; only f2 (the expression for y'') depends on the particular equation. (Key point: Changing the equation) 4. In the RK-4 stage k2/l2, what arguments are passed to f1 and f2? A. x + h/2, y + k2/2, z + l2/2 B. x + h/2, y + k1/2, z + l1/2 C. x + h, y + k1, z + l1 D. x, y + k1/2, z Answer: B. The second stage evaluates at the midpoint using half of the first-stage increments of both y and z. 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