Math Problem Statement

Apply Euler's method to the following system with initial condition (3, -4) and step size Δt = 1/3. Complete the first two steps.

Solution

To solve this problem using Euler's method, let's first understand the system of differential equations provided and the initial condition. Euler's method allows us to approximate solutions for differential equations step by step.

The system of equations:

dxdt=x+y\frac{dx}{dt} = x + y dydt=6y\frac{dy}{dt} = 6y

Initial conditions:

x(0)=3,y(0)=4x(0) = 3, \quad y(0) = -4 The step size is given as Δt=13\Delta t = \frac{1}{3}.

Euler's Method Formula:

For each step, Euler's method updates xx and yy as follows: xn+1=xn+Δtf(xn,yn)x_{n+1} = x_n + \Delta t \cdot f(x_n, y_n) yn+1=yn+Δtg(yn)y_{n+1} = y_n + \Delta t \cdot g(y_n) where f(x,y)=x+yf(x, y) = x + y and g(y)=6yg(y) = 6y are the right-hand sides of the equations.

Step 1: First Iteration (x1x_1, y1y_1)

At t0=0t_0 = 0, we start with: x0=3,y0=4x_0 = 3, \quad y_0 = -4

  1. Compute f(x0,y0)=x0+y0=3+(4)=1f(x_0, y_0) = x_0 + y_0 = 3 + (-4) = -1.

    • Then update x1x_1: x1=x0+Δtf(x0,y0)=3+13(1)=313=2.67x_1 = x_0 + \Delta t \cdot f(x_0, y_0) = 3 + \frac{1}{3} \cdot (-1) = 3 - \frac{1}{3} = 2.67
  2. Compute g(y0)=6y0=6(4)=24g(y_0) = 6y_0 = 6 \cdot (-4) = -24.

    • Then update y1y_1: y1=y0+Δtg(y0)=4+13(24)=48=12y_1 = y_0 + \Delta t \cdot g(y_0) = -4 + \frac{1}{3} \cdot (-24) = -4 - 8 = -12

So, after the first iteration: x1=2.67,y1=12x_1 = 2.67, \quad y_1 = -12

Step 2: Second Iteration (x2x_2, y2y_2)

Now, using x1=2.67x_1 = 2.67 and y1=12y_1 = -12:

  1. Compute f(x1,y1)=x1+y1=2.67+(12)=9.33f(x_1, y_1) = x_1 + y_1 = 2.67 + (-12) = -9.33.

    • Then update x2x_2: x2=x1+Δtf(x1,y1)=2.67+13(9.33)=2.673.11=0.44x_2 = x_1 + \Delta t \cdot f(x_1, y_1) = 2.67 + \frac{1}{3} \cdot (-9.33) = 2.67 - 3.11 = -0.44
  2. Compute g(y1)=6y1=6(12)=72g(y_1) = 6y_1 = 6 \cdot (-12) = -72.

    • Then update y2y_2: y2=y1+Δtg(y1)=12+13(72)=1224=36y_2 = y_1 + \Delta t \cdot g(y_1) = -12 + \frac{1}{3} \cdot (-72) = -12 - 24 = -36

So, after the second iteration: x2=0.44,y2=36x_2 = -0.44, \quad y_2 = -36

Final Answer:

  • x1=2.67x_1 = 2.67
  • y1=12y_1 = -12
  • x2=0.44x_2 = -0.44
  • y2=36y_2 = -36

Would you like more details or have any questions?


Here are 5 related questions:

  1. How does the accuracy of Euler's method depend on the step size Δt\Delta t?
  2. What are the limitations of using Euler’s method for solving differential equations?
  3. How can we improve the accuracy of Euler’s method without changing the step size?
  4. What other numerical methods can be used for solving systems of differential equations?
  5. Can Euler’s method be applied to nonlinear systems of differential equations?

Tip: The smaller the step size Δt\Delta t, the more accurate Euler’s method is, but at the cost of more computations.

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Math Problem Analysis

Mathematical Concepts

Numerical Methods
Differential Equations
Euler's Method

Formulas

x_{n+1} = x_n + Δt * f(x_n, y_n)
y_{n+1} = y_n + Δt * g(y_n)

Theorems

Euler's Method for Ordinary Differential Equations

Suitable Grade Level

Grades 11-12