Backward Euler Method for First Order ODEs in C++ Code | Plot Soln using GNU

Solving y' = -2y using the Implicit (Backward) Euler Method in C++ In this video, I walk through a C++ program that numerically solves the ODE dy/dx = -2y with the initial condition y(0) = 1, using the Implicit Euler method. 🔹 What the code does: Starts at x = 0, y = 1 Uses a step size h = 0.1 up to x = 1 Updates y at each step using the implicit Euler formula: y_(n+1) = y_n / (1 + 2h) Prints the results to the console and saves them to a data file (rd14.dat) for plotting/analysis 🔹 Why Implicit Euler? Unlike the explicit Euler method, the implicit (backward) Euler method is more numerically stable, especially for stiff differential equations — making it a great starting point for anyone learning numerical methods for ODEs. 🔹 Topics covered: Numerical methods in C++ Implicit/Backward Euler method File I/O in C++ (fstream) Solving first-order ODEs numerically 💻 Full code included in the video — feel free to try it yourself and compare with the exact solution y(x) = e^(-2x)! 📌 Like, subscribe, and comment if you liked the video #Cpp #NumericalMethods #ODE #ImplicitEuler #Programming #Coding #EngineeringMath #ScientificComputing