if $\dfrac {dy}{dx}=x+y^2 \space \space given\space y=1$ When $x=0$ in steps of $h=0.1 $

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

Marks : 8

Year : 2013

$$Let f(x,y) \therefore \dfrac {dy}{dx}=x+y^2$$

Part I

$H=0.1 \space x_0 = 0$ and $y_0 =1 $

By Runge kutta method of 4th order

$K_1 = hf(x_0 , y_0) = 0.1 × (0,1)\\ = 0.1 × (0+1^2) = 0.1\\ K_2=hf\Bigg(x_0+\dfrac h2,y_0+\dfrac {k_1}2\Bigg)=0.1\times f(0+\dfrac {0.1}2,1+\dfrac {0.1}2)\\ =0.1\times\Bigg(\dfrac {0.1}2+\Bigg(\dfrac {2.1}2\Bigg)^2\Bigg)\\ =0.1\times (0.05+(1.05)^2)\\ =0.1153 \\ K_3=hf\Bigg(x_0+\dfrac h2,y_0+\dfrac {k_1}2\Bigg) =0.1\times f\Bigg(0+\dfrac {0.1}2,\dfrac {1+0.1153}2\Bigg) \\ =0.1\times\Bigg[\dfrac {0.1}2+\Bigg(\dfrac {2.1153}2\Bigg)^2\Bigg]\\ =0.1169\\ K_4=hf(x_0+h,y_0+k_3)=0.1\times f(0+0.1,1+0.1169)\\ =0.1[0.1+(1.1169)^2] \\ =0.1347\\ \therefore K=\dfrac 16(k_1+2k_2+2k_3+k_4)\\ =\dfrac 16(0.1+0.1153\times 2+0.1169\times 2+ 0.1347)\\ =0.1165\\ \therefore y=y_0+k=1+ 0.1165=1.1165\\ \therefore \text{The correct value of y when}\space x=0-1 \space is\space 1.1165\\ Part II \\ H=0.1 , x_1=0.1 \space \& \space y_1=1.1165 \\ k_1=hf(x_1,y_1)=0.1f(0.1,1.1105) \\ =0.1[0.1+(1.1165)^2]\\ =0.1347 \\ k_2=hf\Big(x_1+\dfrac h2 ,y_1+\dfrac {k_1}2\Big)\\ =0.1\times f\Big(0.1+\dfrac {0.1}2,1.1165+\dfrac {0.1347}2\Big)\\ =0.1[0.15+(1.1838)^2]\\ =0.1551\\ k_3=hf\Big(x_1+\dfrac h2,y_1+\dfrac {k_2}2\Big)\\ =0.1\times f\Big(0.1+\dfrac {0.1}2,1.1165+\dfrac {0.1551}2\Big)\\ =0.1\times [0.15 +(1.1941)^2]\\ =0.1576\\ K_4=hf(x_1+h_1,y_1+k_3)\\ =0.1f(0.1+0.1,1.1165+0.1576)\\ =0.1[0.2+(1.271)^2]\\ =0.1823\\ k=\dfrac 16(k_1+2k_2+2k_3+k_4) \\ =\dfrac 16(0.1347+2\times 0.1551+2\times 0.1576+ 0.1823)\\ =0.1571\\ \therefore y=y_1+k=1.1165+0.1571=1.2736 \\ \therefore \text{The correct value of y when }x=0.2 \space is\space 1.2736 $