Reduce the second order differential equation (given in the image) to a system o

Question

Reduce the second order differential equation (given in the image) to a system of first order differential equations and solve the equations using the FOURTH ORDER Runge-Kutta Method for simultaneous equations, to evaluate y(1.2) taking h=0.2 and 0.1, using the following values:

a = + 10
b = + 3
c = + 30
k1 = + 25

k2 = - 4.5

Second Order Differential Equation:

y" + ay' + by = c

With Initial Conditions:

y(1) = k1, [(y')_x=1_] = k2

Consider the following second order differential equation d2y/dx2 + a dy/dx + by = c with the initial conditions y(1) = kappa 1, (dy/dx)x = 1 = kappa 2. Reduce the second order differential equation to a system of first order differential equations and solve the equations using the FOURTH ORDER Runge-Kutta method for simultaneous equations, to evaluate y(1,2) taking h=0.2 and 0.1, using the following marked data:

Explanation / Answer

y'' + ay' + by = c

y(1) = k1

y'(1) = k2

We have, a = 10, b = 3, c = 30, k1 = 25, k2 = -4.5

Thus,

y'' + 10y' + 3y = 30

y(1) = 25

y'(1) = -4.5

Put y' = z.............(1)

y(1) = 25

Thus, z' + 10z + 3y = 30

z' = 30 - 10z - 3y.................(2)

z(1) = -4.5

We have h = 0.2,

y(1.2) = y(1 + 0.2) = y(1) + (1/6)*(k1,1 + 2k2,1 + 2k3,1 + k4,1)

z(1.2) = z(1 + 0.2) = z(1) + (1/6)*(k1,2 + 2k2,2 + 2k3,2 + k4,2)

where

k1,1 = h*z(1)

k1,2 = -h*1*y(1)

k2,1 = h[z(1) + (1/2)k1,2]

k2,2 = -h*(1 + h/2)[y(1) + (1/2)k1,2]

k3,1 = h*[z(1) + (1/2)*k2,2]

k3,2 = -h*(1 + h/2) [y(1) + (1/2)k2,2]

k4,1 = h*[z(1) + k3,2]

k4,2 = -h*(1 + h)[y(1) + k3,2]

Putting values,

k1,1 = 0.2*(-4.5) = -0.9

k1,2 = -0.2*1*25 = -5

k2,1 = 0.2*[-4.5 + (1/2)*(-5)] = -1.4

k2,2 = -0.2*(1 + 0.2/2)*[25 + (1/2)*(-5)] = -4.95

k3,1 = 0.2*[-4.5 + (1/2)*(-4.95)] = -1.395

k3,2 = -0.2*(1+0.2/2)*[25 + (1/2)*(-4.95)] = -4.9555

k4,1 = 0.2*[-4.5 + (-4.9555)] = -1.8911

k4,2 = -0.2*(1+0.2)[25 + (-4.9555)] = -4.81068

Thus, y(1.2) = 25 + (1/6)[-0.9 + 2*(-1.4) + 2*(-1.395) + (-1.8911)]

y(1.2) = 23.60315

Taking h = 0.1, we get

k1,1 = 0.1*(-4.5) = -0.45

k1,2 = -0.1*1*25 = -2.5

k2,1 = 0.1*[-4.5 + (1/2)*(-2.5)] = -0.575

k2,2 = -0.1*(1+0.1/2)[25 + (1/2)*(-2.5)] = -2.49375

k3,1 = 0.1*[-4.5 + (1/2)*(-2.49375)] = -0.5746875

k3,2 = -0.1*(1+0.1/2)[25 + (1/2)*(-2.49375)] = -2.494078125

k4,1 = 0.1*[-4.5 + (-2.494078125)] = -0.6994078125

k4,2 = -0.1*(1+0.1)*[25 + (-2.494078125)] = -2.47565140625

y(1.1) = 25 + (1/6)*[-0.45 + 2*(-0.575) + 2*(-0.5746875) + (-2.4765140625)]

y(1.1) = 24.129162265625

z(1.1) = -4.5 + (1/6)*[-2.5 + 2*(-2.49375) + 2*(-2.494078125) + (-2.47565140625)]

z(1.1) = -6.991884609375

Next iteration, y(1.2) = y(1.1 + 0.1)

k1,1 = 0.1*(-6.991884609375) = -0.6991884609375

k1,2 = -0.1*1.1*24.129162265625 = -2.65420784921875

k2,1 = 0.1*[-6.991884609375 + (1/2)*(-2.65420784921875)] = -0.8318988533984375

k2,2 = -0.1*(1.1+0.1/2)[24.129162265625 + (1/2)*(-2.65420784921875)] = -2.622236709216796875

k3,1 = 0.1*[-6.991884609375 + (1/2)*(-2.622236709216796875)] = -0.83030029639833984375

k3,2 = -0.1*(1.1 + 0.1/2)*[24.129162265625 + (1/2)*(-2.622236709216796875)] = -2.6240750497669091796875

k4,1 = 0.1*[-6.991884609375 + (-2.6240750497669091796875)] = -0.96159596591419091796875

y(1.2) = y(1.1 + 0.1) = 24.129162265625 + (1/6)*[-0.6991884609375 + 2*(-0.8318988533984375) + 2*(-0.83030029639833984375) + (-0.96159596591419091796875)]

y(1.2) = 23.298298477884125732421875

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