Question

Consider the initial value problem \(y^{\prime}(t)=t^{2}-3 y^{2}, y(3)=1\) What is the approximation to \(y(3.1)\) given by Euler's method with a time step of \(\Delta t=0.1 ?\)

Short answer

Answer: The approximate value of y(3.1) using Euler's method with a time step of 0.1 is 1.6.

Step-by-step solution

Understand Euler's method formula

Euler's method is used for approximating the solution of an initial value problem of the form \(y^{\prime}(t) = f(t, y(t))\) and an initial condition \(y(t_0) = y_0\), where t is the independent variable, and y is the dependent variable. The method advances the solution from one step to the next as follows: \(y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n)\) In our problem, we have the initial condition \(y(3)=1\), the differential equation \(y^{\prime}(t)=t^{2}-3y^{2}\), and the time step \(\Delta t=0.1\).

Set up the first iteration

In the first iteration of the Euler's method, we need to find the value of \(y(3 + \Delta t)=y(3.1)\) using the initial condition \(y(3)=1\) and the given differential equation \(y^{\prime}(t)=t^{2}-3y^{2}\). We will use the formula: \(y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n)\) where \(y_n=y(3)=1\), \(t_n=3\), \(f(t_n, y_n)=f(3,1)=3^{2}-3(1)^{2}=9-3\), and \(\Delta t=0.1\).

Calculate y(3.1) using Euler's method

Using the formula and the values from Step 2, we can now calculate the approximate value of y(3.1): \(y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n)\) \(y(3.1) = y(3) + \Delta t \cdot f(3,1)\) \(y(3.1) = 1 + 0.1 \cdot (9-3)\) \(y(3.1) = 1 + 0.1 \cdot 6\) Now, we can calculate the value of y(3.1): \(y(3.1) = 1 + 0.6\) \(y(3.1) = 1.6\) The approximation of y(3.1) using Euler's method with a time step of \(\Delta t=0.1\) is 1.6.