Question

If the numerical solution of the initial value problem \[ y' = \frac{t^2}{t + y^3}, y(0) = 1, \] is obtained by the Euler's method with step size of 0.2, then the value of \( y(0.4) \), (rounded off to two decimal places), is \(\underline{\hspace{2cm}}\).

Hint:

Euler's method approximates the solution to differential equations by using the slope at each step to estimate the next value. Use a small step size for better accuracy.

Answer 1

Correct Answer: 0.98

The given differential equation is \( y' = \frac{t^2}{t + y^3} \), with the initial condition \( y(0) = 1 \). We are asked to solve it numerically using Euler's method with a step size of 0.2.

Step 1: Apply Euler's method.
Euler's method for solving ordinary differential equations is given by the formula: \[ y_{n+1} = y_n + h f(t_n, y_n), \] where \( h \) is the step size and \( f(t_n, y_n) \) is the derivative at the point \( (t_n, y_n) \). For this problem, the function \( f(t, y) \) is: \[ f(t, y) = \frac{t^2}{t + y^3}. \] We are given that \( y(0) = 1 \), and we need to compute \( y(0.4) \).

Step 2: Compute \( y(0.2) \).
First, apply Euler's method with \( t_0 = 0 \) and \( y_0 = 1 \): \[ f(0, 1) = \frac{0^2}{0 + 1^3} = 0, \] \[ y_1 = y_0 + h f(t_0, y_0) = 1 + 0.2 \times 0 = 1. \] Thus, \( y(0.2) = 1 \).

Step 3: Compute \( y(0.4) \).
Now, apply Euler's method to compute \( y(0.4) \) using \( y(0.2) = 1 \): \[ f(0.2, 1) = \frac{0.2^2}{0.2 + 1^3} = \frac{0.04}{1.2} \approx 0.03333, \] \[ y_2 = y_1 + h f(t_1, y_1) = 1 + 0.2 \times 0.03333 = 1.00667. \] Thus, \( y(0.4) \approx 1.01 \).