Question

Apply Rungee-Kutta Method of fourth order to find an approximate value of y when x=0.2 given that `(dy)/(dx)=x+y` when y=1 at x=0 with step size h=0.2.

Answer 1

`(dy)/(dx)=x+y`  `x_0=0, y_0=1`, h= 0.2

f(x,y)= x + y

`k_1=h.f(x_0.y_0)=0.2f(0,1)=0.2`

`k_2=h.f(x_0+h/2,y_0+k_1/2)=0.2f(0.1,1.1)=0.24`

`k_3=h.f(x_0+h/2,y_0+k_2/2)=0.2f(0.1,1.12)=0.244`

`k_4=h.f(x_0+h,y_0+k_3)=0.2f(0.2,1.244)=0.2888`

`k=(k_1+2k_2+2k_3+k_4)/6=(0.24+0.48+0.488+0.2888)/6=0.2428`

The value of y at x=0.2 is given by,

y(0.2) = `y_0+k`= 1+ 0.2428
y(0.2) = 1.2428