Eulers method for a first order IVP y fx y yx0 y0 is the f

Euler\'s method for a first order IVP y\' = f(x, y), y(x_0) = y_0 is the following algorithm. From (x_0, y_0) we define a sequence of approximations to the solution of the differential equation so that at the nth stage, we have x_n = x_n-1 + h, y_n = y_n-1 + h middot f(x_n - 1, y_n - 1). In this exercise we consider the IVP y\' = sin(y) with (-0.5) = 2. Use Euler\'s method with h = 0.3 to approximate the solution of the differential equation.

Solution

f(x,y) = siny

x₀ = -0.5, y₀ = 2

f(x₀, y₀) = sin(2) = 0.9093

x₁ = x₀ + h = -0.5 + 0.3 = -0.2

y₁ = y₀ + hf(x₀,y₀)

Also y\' = siny

=> dy /siny = dx

=> cscydy = dx

=> -ln|cscy+coty| = x+C

At x₀ = -0.5, y₀ = 2

=> -ln|csc2 + cot2| = -0.5+C

=> -ln|1.0997-0.4576| = -0.5+C

=> 0.9430 = C

Hence -ln|cscy+coty| = x+0.9430

x₁ = -0.5+0.3 = -0.2

y₁ = 2+(0.3)(0.9093) = 2.27279

Exact solution at (x₁,y₁)

-ln|cscy +coty| = -0.2+0.9430 =0 .743

=> cscy+coty = e^-0.743 = 0.4757

=> y = 2.2536

Error = |2.2536-2.27279| = 0.01919

f(x₁,y₁) = sin(2,27279) = 0.7635

x₂ = -0.2+0.3 = 0.1

y₂ = 2.27279 + (0.3)(0.7635) = 2.50184

Exact solution at x₂ = 0.1

-ln|cscy+coty| = 0.1 + 0.9430 = 1.043

=> cscy+coty = e^-1.043 = 0.3524

=> y = 2.464

Error =|2.464-2.50184| = 0.03784

f(x₂,y₂) = sin(2.50184) = 0.5970

x₃ = 0.1 + 0.3 = 0.4

y₃ = 2.50184 + (0.3)(0.5970) = 2.68094

Exact solution at x₃ = 0.4

-ln|cscy+coty| = 0.4+0.9430 = 1.343

=> cscy+coty = e^-1.343 = 0.2611

=> y = 2.6308

Error = |2.6308-2.68094| = 0.05014

f(x₃,y₃) = sin(2.68094) = 0.4445

x₄ = 0.4+0.3 = 0.7

y₄ = 2.68094 + (0.3)(0.4445) = 2.81429

Exact solution at x₄ = 0.7

-ln|cscy+coty| = 0.7+0.9430 = 1.643

=> cscy+coty = e^-1.643 = 0.1934

=> y = 2.7596

Error = |2.7596-2.81429| = 0.05469

The table of values is iven by :

n=	0	1	2	3	4
xn	-0.5	-0.2	0.1	0.4	0.7
yn	2	2.27279	2.50184	2.68094	2.81429
en		0.01919	0.03784	0.05014	0.05469