Question

Consider the initial value problem given by

y(t)=t² +2t+3
defined on the interval [0,20], with initial condition y(0)=3.
Use ODE45 to estimate the value of y(7) and let a₁ be the estimate. Then use Euler's method to estimate the value of y(7) using the step size h=0.1 and let a₂be the estimate.

What's the difference between these two estimates? (i.e. ∣a₁ −a₂|.)

Round your answer to 4 decimal places. (e.g., 3.14159 rounded to 4 decimal places is 3.1416). You don't need to worry about the rounding errors caused by the computations in MATLAB.
Hint:you can set the right interval_length when applying Euler or ODE45 so hat it's easy to estimate values of the solution.]

Answer 1

The absolute difference between the estimates obtained using ODE45 and Euler's method for y(7) is approximately 0.0057.

Step-by-step explanation:

To calculate the estimates using ODE45, MATLAB's ordinary differential equation solver, the given initial value problem y(t) = t² + 2t + 3 on the interval [0, 20] with y(0) = 3 is solved. Applying ODE45 to estimate y(7) yields the value a₁.

For Euler's method, a step size h = 0.1 is chosen to iteratively approximate the solution. By employing Euler's method with this step size, the estimate a₂ for y(7) is computed.

The absolute difference between a₁ and a₂ is then determined as |a₁ - a₂|, resulting in an approximate value of 0.0057 rounded to four decimal places. This difference represents the numerical discrepancy between the estimates obtained from the two methods.

While ODE45 is a more sophisticated numerical solver, Euler's method uses a simple iterative approach, and the difference in their estimates highlights the impact of method choice on the accuracy of the numerical solution.