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68 lines
2.3 KiB
C++
68 lines
2.3 KiB
C++
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/*----------- RUNG KUTTA METHOD TO SOLVE DIFFERENTIAL EQUATION --------*/
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/* THIS PROGRAM CALCULATES THE VALUE y AT GIVEN VALUE OF x
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USING FOURTH ORDER RUNG KUTTA METHOD. THE FUNCTION y' = f(x,y)
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IS DEFINED IN THE PROGRAM.
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y' = 1 + y*y
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Hence f(x,y) = 1 + y*y
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INPUTS : 1) Initial values of x and y.
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2) Step size h.
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OUTPUTS : Calculated values of y at every step. */
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/*------------------------------ PROGRAM --------------------------*/
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#include<stdio.h>
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#include<math.h>
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#include<stdlib.h>
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#include<conio.h>
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void main()
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{
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double f (double x0,double y0); /* DECLARATION OF A FUNCTION f */
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double y0,y1,x0,x1,h,x,k1,k2,k3,k4,k;
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int i,n;
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clrscr();
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printf("\n\tRUNG KUTTA METHOD TO SOLVE DIFFERENTIAL EQUATION\n");
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printf("\n\nEnter x0 = ");
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scanf("%lf",&x0); /* ENTER VALUE OF x0 */
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printf("\n\nEnter y0 = ");
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scanf("%lf",&y0); /* ENTER VALUE OF y0 */
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printf("\n\nEnter the value of x at which y is to be found = ");
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scanf("%lf",&x); /* ENTER VALUE OF x */
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printf("\n\nEnter the value of h = ");
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scanf("%lf",&h); /* ENTER THE VALUE OF h */
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i = 0;
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printf("\nPress any key to see step by step display of results...\n");
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while(x0 < x) /* LOOP TO CALCULATE y USING RUNG KUTTA METHOD */
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{
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i++;
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k1 = f(x0,y0);
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k2 = f(x0+h/2, y0+(h*k1/2));
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k3 = f(x0+h/2, y0+(h*k2/2));
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k4 = f(x0+h, y0+h*k3);
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/* CALCULATION OF k USING RUNG KUTTA METHOD */
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y1 = y0 + (h/6)*(k1 + 2*k2 + 2*k3 + k4);
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/* CALCULATION OF y FROM VALUES OF k */
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x1 = x0 + h;
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printf("\nx%d = %lf y%d = %lf",i,x1,i,y1);
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x0 = x1;
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y0 = y1;
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getch();
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}
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}
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/*---------------------------------------------------------------------*/
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double f ( double x,double y) /* FUNCTION TO CALCULATE VALUE OF f(x,y)*/
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{
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double y_dash;
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y_dash = 1 + y*y; /* function f(x,y) = y' = 1 + y*y */
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return(y_dash);
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}
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/*------------------------ END OF PROGRAM -----------------------------*/
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