Problem 3 25 pts Solve the following differential equation u

Problem 3 (25 pts.) Solve the following differential equation using Euler\'s method over the interval from x-otox-1 with a step size h 0.5 and h 0.25. Consider and initial value of y 1 atx o Also, calculate the absolute relative error of your approximations only at x-1, for each time step considering that the analytical solution is obtained employing the following equation 1.63733 General form of Euler\'s Method: New value old value slope x step size ton 0.5 632 3 O + 1 51 r)07322243

Solution

Given, x₀ = 0 and y₀ = 1

and let f(x,y) = slope = y\' = yx² - y

For Step Size (h) of 0.5

f₀ = f(0,1) = 1 (0)² - 1 = -1

y₁ = y₀ + h * f₀ = 1 + 0.5 (-1) = 0.5

and x₁ = x₀ + h = 0.5

Now f₁ = f(0.5, 0.5) = 0.5 (0.5)² - 0.5 = -0.375

y₂ = y₁ + h * f₁ = 0.5 + 0.5 (-0.375) = 0.3125

and x₂ = x₁ + h = 1

Thus for x₂ = 1, y₂ = 0.3125

Also, from given analytical solution, y(1) = e^{(1/3 - 1)} = 0.513417119

Absolute error = |0.513417119 - 0.3125| = 0.200917119

For Step Size (h) of 0.25

f₀ = f(0,1) = 1 (0)² - 1 = -1

y₁ = y₀ + h * f₀ = 1 + 0.25 (-1) = 0.75

and x₁ = x₀ + h = 0.25

Now f₁ = f(0.25, 0.75) = 0.75 (0.25)² - 0.75 = -0.70313

y₂ = y₁ + h * f₁ = 0.75 + 0.25 (-0.70313) = 0.574219

and x₂ = x₁ + h = 0.5

Now f₂ = f(0.5, 0.574219) = -0.43066

y₃ = y₂ + h * f₂ = 0.574219 + 0.25 (-0.43066) = 0.466553

and x₃ = x₂ + h = 0.75

Now f₃ = f(0.75, 0.466553) = -0.20412

y₄ = y₃ + h * f₃ = 0.466553 + 0.25 (-0.20412) = 0.415524

and x₄ = x₃ + h = 1

Thus for x₂ = 1, y₂ = 0.415524

Also, from given analytical solution, y(1) = e^{(1/3 - 1)} = 0.513417119

Absolute error = |0.513417119 - 0.415524| = 0.09789358998