/******************************************************************************
Solution using Backward Euler
*******************************************************************************/
// C++ code for solving the differential equation
// using Predictor-Corrector or Modified-Euler method
// with the given conditions, y(0) = 0.5, step size(h) = 0.2
// to find y(1)
#include <bits/stdc++.h>
using namespace std;
// consider the differential equation
// for a given x and y, return v
double f(double x, double y)
{
double v = y - 2 * x * x + 1;
return v;
}
// predicts the next value for a given (x, y)
// and step size h using Euler method
double predict(double x, double y, double h)
{
// value of next y(predicted) is returned
double y1p = y + h * f(x, y);
return y1p;
}
// corrects the predicted value
// using Modified Euler method
double correct(double x, double y,
double x1, double y1,
double h)
{
// (x, y) are of previous step
// and x1 is the increased x for next step
// and y1 is predicted y for next step
double e = 0.00001;
double y1c = y1;
do {
y1 = y1c;
y1c = y + 0.5 * h * (f(x, y) + f(x1, y1));
} while (fabs(y1c - y1) > e);
// every iteration is correcting the value
// of y using average slope
return y1c;
}
void printFinalValues(double x, double xn,
double y, double h)
{
while (x < xn) {
double x1 = x + h;
double y1p = predict(x, y, h);
double y1c = correct(x, y, x1, y1p, h);
x = x1;
y = y1c;
}
// at every iteration first the value
// of for next step is first predicted
// and then corrected.
cout << "The final value of y at x = "
<< x << " is : " << y << endl;
}
int main()
{
// here x and y are the initial
// given condition, so x=0 and y=0.5
double x = 0, y = 0.5;
// final value of x for which y is needed
double xn = 1;
// step size
double h = 0.2;
printFinalValues(x, xn, y, h);
return 0;
}
