Modify a C code to find the solution of the differential equ
Solution
// C program to implement Runge Kutta method
 #include<stdio.h>
 
 // A differential equation \"dy/dx = 1/1-y*y*y+0.3*x*x\"
 float dydx(float x, float y)
 {
     return(1/(1-y*y*y+0.3*x*x));
 }
 
 // Finds value of y for a given x using step size h
 // and initial value y0 at x0.
 float rungeKutta(float x0, float y0, float x, float h)
 {
     // Count number of iterations using step size or
     // step height h
     int n = (int)((x - x0) / h);
 
     float k1, k2, k3, k4, k5;
 
     // Iterate for number of iterations
     float y = y0;
     for (int i=1; i<=n; i++)
     {
         // Apply Runge Kutta Formulas to find
         // next value of y
         k1 = h*dydx(x0, y);
         k2 = h*dydx(x0 + 0.5*h, y + 0.5*k1);
         k3 = h*dydx(x0 + 0.5*h, y + 0.5*k2);
         k4 = h*dydx(x0 + h, y + k3);
 
         // Update next value of y
         y = y + (1.0/6.0)*(k1 + 2*k2 + 2*k3 + k4);;
         printf(\"\ \  y(%.4lf) = %.3lf \",x0+h,y);
 
         // Update next value of x
         x0 = x0 + h;
     }
 
     return y;
 }
 
 // Driver method
 int main()
 {
     float x0 = 2, y = 3, x = 3, h = 0.1;
     printf(\"\ The value of y at x is : %f\",
             rungeKutta(x0, y, x, h));
     return 0;
 }
 OUTPUT:
y(2.1000) = 2.996
y(2.2000) = 2.992
y(2.3000) = 2.988
y(2.4000) = 2.984
y(2.5000) = 2.979
y(2.6000) = 2.975
y(2.7000) = 2.971
y(2.8000) = 2.966
y(2.9000) = 2.962
y(3.0000) = 2.957
The value of y at x is : 2.957478sh-4.3$

