Question

Consider the differential equation \(\frac{dy}{dx} = x + y\) with the initial condition \(y(0) = 1\). Using the modified Euler's method, the second approximation to \(y(h)\), where \(h = 0.05\) (step size), is equal to ______. (rounded off to TWO decimal places)

Hint:

First find the predictor y1 = y0 + h f(x0,y0), then apply the corrector formula twice to get the second approximation.

Answer 1

Correct Answer: 1.05

The modified Euler's method comes from approximating the integral form of the differential equation using the trapezoidal rule, and then repeating the correction step to refine the estimate. We can lay out the whole calculation in a short table instead of a step by step narrative.

1. Why the formula looks the way it does: Integrating $\frac{dy}{dx}=f(x,y)$ from $x_0$ to $x_1=x_0+h$ gives $y(x_1) = y_0 + \int_{x_0}^{x_1} f(x,y)\,dx$. Approximating this integral by the trapezoidal rule (averaging the slope at the two ends) gives $y_1 \approx y_0 + \frac{h}{2}[f(x_0,y_0) + f(x_1,y_1)]$. Since $y_1$ appears on both sides, we solve it by repeated substitution, starting from a first guess.
2. Round 0 (initial guess by plain Euler): $y_1^{(0)} = y_0 + h f(x_0,y_0) = 1 + 0.05(0+1) = 1.05$.
3. Round 1 (first correction): $f(x_1, y_1^{(0)}) = 0.05 + 1.05 = 1.10$, so $y_1^{(1)} = 1 + \frac{0.05}{2}(1 + 1.10) = 1 + 0.025(2.10) = 1.0525$.
4. Round 2 (second correction): $f(x_1, y_1^{(1)}) = 0.05 + 1.0525 = 1.1025$, so $y_1^{(2)} = 1 + \frac{0.05}{2}(1 + 1.1025) = 1 + 0.025(2.1025) = 1.0525625$.

The value $y_1^{(2)} = 1.0525625$ is the second approximation asked for in the question.

Let's summarize:

• The predictor gives a rough first guess using the plain Euler formula.
• Each pass of the corrector formula (based on the trapezoidal rule) refines this guess, and the second approximation means the corrector has been applied twice.

Rounding $1.0525625$ to two decimal places gives $1.05$.

$$\boxed{1.05}$$