Math Problem Statement

Use​ Euler's method to approximate the solution to the given initial value problem at the points xequals0.1​, 0.2​, 0.3​, 0.4​, and 0.5​, using steps of size 0.1 ​(hequals0.1​). StartFraction dy Over dx EndFraction equalsy left parenthesis 4 minus y right parenthesis​, ​y(0​)equals8 Question content area bottom Part 1 The approximate solution to StartFraction dy Over dx EndFraction equalsy left parenthesis 4 minus y right parenthesis​, ​y(0​)equals8​, at the point xequals0.1 is    enter your response here.

Solution

We will use Euler's method to approximate the solution to the differential equation:

dydx=y(4y),y(0)=8\frac{dy}{dx} = y(4 - y), \quad y(0) = 8

with step size h=0.1h = 0.1, at x=0.1,0.2,0.3,0.4,x = 0.1, 0.2, 0.3, 0.4, and 0.50.5.

Formula for Euler's Method:

yn+1=yn+hf(xn,yn)y_{n+1} = y_n + h \cdot f(x_n, y_n) where f(x,y)=y(4y)f(x, y) = y(4 - y).

Step 1: Initialize

At x0=0x_0 = 0, y0=8y_0 = 8.

Step-by-Step Calculation

  1. At x0=0,y0=8x_0 = 0, y_0 = 8: f(0,8)=8(48)=8(4)=32f(0, 8) = 8(4 - 8) = 8(-4) = -32 y1=y0+hf(0,8)=8+0.1(32)=83.2=4.8y_1 = y_0 + h \cdot f(0, 8) = 8 + 0.1(-32) = 8 - 3.2 = 4.8

  2. At x1=0.1,y1=4.8x_1 = 0.1, y_1 = 4.8: f(0.1,4.8)=4.8(44.8)=4.8(0.8)=3.84f(0.1, 4.8) = 4.8(4 - 4.8) = 4.8(-0.8) = -3.84 y2=y1+hf(0.1,4.8)=4.8+0.1(3.84)=4.80.384=4.416y_2 = y_1 + h \cdot f(0.1, 4.8) = 4.8 + 0.1(-3.84) = 4.8 - 0.384 = 4.416

  3. At x2=0.2,y2=4.416x_2 = 0.2, y_2 = 4.416: f(0.2,4.416)=4.416(44.416)=4.416(0.416)=1.837056f(0.2, 4.416) = 4.416(4 - 4.416) = 4.416(-0.416) = -1.837056 y3=y2+hf(0.2,4.416)=4.416+0.1(1.837056)=4.4160.1837056=4.2322944y_3 = y_2 + h \cdot f(0.2, 4.416) = 4.416 + 0.1(-1.837056) = 4.416 - 0.1837056 = 4.2322944

  4. At x3=0.3,y3=4.2322944x_3 = 0.3, y_3 = 4.2322944: f(0.3,4.2322944)=4.2322944(44.2322944)=4.2322944(0.2322944)0.98297f(0.3, 4.2322944) = 4.2322944(4 - 4.2322944) = 4.2322944(-0.2322944) \approx -0.98297 y4=y3+hf(0.3,4.2322944)=4.2322944+0.1(0.98297)=4.23229440.0982974.134y_4 = y_3 + h \cdot f(0.3, 4.2322944) = 4.2322944 + 0.1(-0.98297) = 4.2322944 - 0.098297 \approx 4.134

  5. At x4=0.4,y4=4.134x_4 = 0.4, y_4 = 4.134: f(0.4,4.134)=4.134(44.134)=4.134(0.134)0.554f(0.4, 4.134) = 4.134(4 - 4.134) = 4.134(-0.134) \approx -0.554 y5=y4+hf(0.4,4.134)=4.134+0.1(0.554)4.078y_5 = y_4 + h \cdot f(0.4, 4.134) = 4.134 + 0.1(-0.554) \approx 4.078

Final Results

The approximate values of y(x)y(x) are:

  • y(0.1)=4.8y(0.1) = 4.8
  • y(0.2)4.416y(0.2) \approx 4.416
  • y(0.3)4.232y(0.3) \approx 4.232
  • y(0.4)4.134y(0.4) \approx 4.134
  • y(0.5)4.078y(0.5) \approx 4.078

Let me know if you need the calculations detailed further or for any clarifications!


  1. What is Euler's method and how does it approximate solutions?
  2. Why is the step size important in Euler's method?
  3. How does the accuracy of the method change with smaller step sizes?
  4. What are the limitations of Euler's method?
  5. Can Euler's method be applied to non-linear differential equations?

Tip: Always cross-check with smaller step sizes for higher accuracy when using Euler’s method.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Numerical Methods
Euler's Method

Formulas

y_{n+1} = y_n + h * f(x_n, y_n)
f(x, y) = y(4 - y)

Theorems

Euler's Method for Numerical Approximation

Suitable Grade Level

Undergraduate Mathematics or Advanced High School Calculus