×

# Doubt in Eulers method with smaller step size

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?

Note by Romil Punetha
1 year, 7 months ago

Sort by:

class euler { public static void main(String args[]) { float i,fn=5,d,step,f7; for(i=1;i<=12;i=i+1/3) { d=iii; 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 · 1 year, 7 months ago

while(x<=limit) { diff=xxx; 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. · 1 year, 7 months ago