Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 8

Year : 2013

$$Let \space f(x,y)= \dfrac {dy}{dx}=-y-xy^2$$

Part I:

Here $h=0.1 , x_0 = 0 \space \&\space y_0 = 1$

By Runge-kutta method of fourth order.

$K_1 = hf(x_0 , y_0)$

$= 0.1 f(0,1)$

$= 0.1 × (-1- (0 × 1)^2)$

$= -0.1\\ K_2=hf\Big(x_0+\dfrac h2,y_0+\dfrac {k_1}2\Big)\\ 0.1f\Big(0+\dfrac {0.1}2,1-\dfrac {0.1}2\Big)\\ =0.1f(0.05,0.95)\\ =0.1\times (-0.95-0.05\times(0.95)^2)\\ =-0.09951\\ K_3=hf\Big(x_0+\dfrac h2,y_0+\dfrac {k_2}2\Big)\\ =0.1f\Big(0+\dfrac {0.1}2,1-\dfrac {0.09951}2\Big)\\ =0.1f(0.05,0.9502)\\ =0.1(-0.9502-0.1\times 0.9502)\\ =-0.09954\\ K_4=h.f(x_0+h,y_0+h)\\ =0.1f((0+0.1),(1-0.09954))\\ =0.1(-0.9005-0.1\times 0.9005^2)\\ =-0.09815\\ K=\dfrac 16(k_1+2k_2+2k_3+k_4)\\ =\dfrac 16(-0.1-2\times0.09951-2\times 0.09954-0.09815)\\ =-0.09938\\ \therefore y=y_0+k=1-0.09938=0.90062 \\ \text{The correct value of y when x is } 0.1 \space is \space 0.9006.$