I have a doubt in applying eulers method with smaller step size. While I got all the problems with step size =1 correct, I am unable to get any in the other category correct. Isnt eulers method this?

\({ y }_{ n+1 }={ y }_{ n }+\quad (step\quad size\quad *\quad dy/dx)\) ?

Im trying probs like dy/dx= x^3 , f(1)=5, step size = 1/3, find f(12)- f(7). I even wrote a program to do this but still im getting the wrong answer. Can someone please help?

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TopNewestclass euler { public static void main(String args[]) { float i,fn=5,d,step,f7; for(i=1;i<=12;i=i+1/3) { d=i

ii; step=1/3;fn=fn+step*d; if(i==7) { f7=fn; }

} System.out.println(fn - f7); } } ye try kr .. P.S: n compiler here :D – Mayank Uniyal · 2 years, 2 months ago

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xx; printf("%f \t %f \t %f \n",x,y,diff); y=y+ (step*diff); x=x+step;Guess ive used the same thing. Check the screenshot, its from the results of the same code. – Romil Punetha · 2 years, 2 months ago

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