Math Problem Statement

Solve the differential equation by Euler’s method to solve the initial

value problem over the interval x = 0 to 2 with h = 0.5 where

2 1.1 dy yx y dx = - ; where y(0) = 1.

Solution

Euler's method is a numerical technique used to solve ordinary differential equations (ODEs) with a given initial value. The basic idea is to approximate the solution by stepping through the interval in small increments, updating the value of the function at each step.

Given the differential equation:

dydx=21.1xy\frac{dy}{dx} = 2 - 1.1xy

with the initial condition y(0)=1y(0) = 1, and we are to solve it over the interval x=0x = 0 to x=2x = 2 with a step size h=0.5h = 0.5.

Step-by-Step Solution Using Euler's Method:

  1. Initial Condition: x0=0,y0=1x_0 = 0, \, y_0 = 1

  2. Recurrence 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)=21.1xyf(x, y) = 2 - 1.1xy.

  3. Step 1: Calculate y1y_1 at x1=0.5x_1 = 0.5: f(x0,y0)=21.101=2f(x_0, y_0) = 2 - 1.1 \cdot 0 \cdot 1 = 2 y1=y0+hf(x0,y0)=1+0.52=1+1=2y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.5 \cdot 2 = 1 + 1 = 2

  4. Step 2: Calculate y2y_2 at x2=1.0x_2 = 1.0: f(x1,y1)=21.10.52=21.1=0.9f(x_1, y_1) = 2 - 1.1 \cdot 0.5 \cdot 2 = 2 - 1.1 = 0.9 y2=y1+hf(x1,y1)=2+0.50.9=2+0.45=2.45y_2 = y_1 + h \cdot f(x_1, y_1) = 2 + 0.5 \cdot 0.9 = 2 + 0.45 = 2.45

  5. Step 3: Calculate y3y_3 at x3=1.5x_3 = 1.5: f(x2,y2)=21.11.02.45=22.695=0.695f(x_2, y_2) = 2 - 1.1 \cdot 1.0 \cdot 2.45 = 2 - 2.695 = -0.695 y3=y2+hf(x2,y2)=2.45+0.5(0.695)=2.450.3475=2.1025y_3 = y_2 + h \cdot f(x_2, y_2) = 2.45 + 0.5 \cdot (-0.695) = 2.45 - 0.3475 = 2.1025

  6. Step 4: Calculate y4y_4 at x4=2.0x_4 = 2.0: f(x3,y3)=21.11.52.1025=23.46575=1.46575f(x_3, y_3) = 2 - 1.1 \cdot 1.5 \cdot 2.1025 = 2 - 3.46575 = -1.46575 y4=y3+hf(x3,y3)=2.1025+0.5(1.46575)=2.10250.732875=1.369625y_4 = y_3 + h \cdot f(x_3, y_3) = 2.1025 + 0.5 \cdot (-1.46575) = 2.1025 - 0.732875 = 1.369625

Final Answer:

The approximate values of y(x)y(x) at the points x=0.5,1.0,1.5,2.0x = 0.5, 1.0, 1.5, 2.0 using Euler's method with step size h=0.5h = 0.5 are:

y(0.5) & = 2, \\ y(1.0) & = 2.45, \\ y(1.5) & = 2.1025, \\ y(2.0) & = 1.369625. \end{aligned}$$ Would you like more details or have any other questions? --- ### Expanding Understanding: 1. How does Euler's method compare to other numerical methods like the Runge-Kutta method in terms of accuracy? 2. What happens if we decrease the step size $$h$$? How does it affect the accuracy of the solution? 3. Can you describe the general approach to solving differential equations analytically versus numerically? 4. What are the limitations of Euler's method, and how can they be mitigated? 5. How would you extend Euler’s method to solve a system of differential equations? ### Tip: When using Euler's method, the smaller the step size $$h$$, the more accurate the approximation will be, but it will require more computations.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Euler's Method
Numerical Methods

Formulas

Euler's Method: y_{n+1} = y_n + h \cdot f(x_n, y_n)
Differential Equation: \frac{dy}{dx} = 2 - 1.1xy

Theorems

-

Suitable Grade Level

Advanced Undergraduate