An autonomous system of two first order differential equations can be written as

Question

An autonomous system of two first order differential equations can be written as: du/dt = f(u + v), u(t_0) = u_0, du/dt = g(u, v), u(t_0) = v_0. A third order explicit Runge-Kutta scheme for an autonomous system of two first order equations is k_1 = hf(u_n, u_n), l_1 = hg(u_n, u_n) k_2 = hf(u_n + 1/3 k_1, u_n + 1/3l_1) l_2 = hg (u_n + 1/3 k_1, u_n + 1/3 l_1), k_3 = hf (u_n + 2/3 k_2, u_n 2/3 l_2_), l_3 = hg (u_n + 2/3 k_2, u_n + 2/3 l_2), u_n + 1 = u_n + 1/4 (k_1 + 3k_3), u_n + 1 = u_n + 1/4 (l_1 + 3l_3). Consider the following second order differential equation, d^2y/dt^2 - 2 dy/dt - 5y^2 = 0.8, with y(1)=0 and y"(1) = 4. Use the Runge-Kutta scheme to find an approximate solutions of the second order differential equation, at t = 1.2, if the step size h = 0.1. Maintain at least eight decimal digit accuracy throughout all your calculations. You may express your answer as a five decimal digit number, for example 16.17423. YOU DO NOT HAVE TO ROUND YOUR FINAL ESTIMATE. y(1.2) almostequalto

Explanation / Answer

let dy/dt = x

then

dx/dt -2x-5y^2=0.8

dx/dt=2x+5y^2+0.8 =g(x,y) and dy/dt =x = f(x,y)

these are the two simultaneously DE have to solve using RK method

y(1)=0 and x(1) = 4

h=0.1

k1=0.1* x(1)=0.4 l1= 0.1*( 2x(1)+5y(1)^2+0.8=0.88

k2=0.1*f(x(1)+k1/3,y(1)+l1/3) = 0.41333333 l2=0.1*g(x(1)+k1/3,y(1)+l1/3) =  0.949688888888

k3=0.1*f(x(1)+2k1/3,y(1)+2l1/3)= 0.426666666666 l3=0.1*g(x(1)+2k1/3,y(1)+2l1/3) =  1.1054222222

now

y(1.1)=y(1)+1/4(k1+3k3)= 0.4200 x(1.1)=x(1)+1/4(l1+3l3)=  5.049066666

itration 2

now assume

k1=0.1* x(1.1)= 0.5049066666 l1= 0.1*( 2x(1.1)+5y(1.1)^2+0.8=  1.1780133332

k2=0.1*f(x(1.1)+k1/3,y(1.1)+l1/3) = 0.52173688882 l2=0.1*g(x(1.1)+k1/3,y(1.1)+l1/3) =   1.4536909450

k3=0.1*f(x(1.1)+2k1/3,y(1.1)+2l1/3)= 0.5385671110 l3=0.1*g(x(1.1)+2k1/3,y(1.1)+2l1/3) = 1.883559158

now

y(1.2)=y(1.1)+1/4(k1+3k3)=  0.950151999 x(1.2)=x(1.1)+1/4(l1+3l3)= 6.756239368

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