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

Step 1: Set up the known values.
We have \(f(x,y) = x + y\), starting point \(x_0 = 0\), \(y_0 = 1\), and step size \(h = 0.05\).
We want to estimate \(y\) at \(x_1 = x_0 + h = 0.05\).

Step 2: Find the first (predictor) estimate.
The modified Euler's method first uses the plain Euler formula to get a rough starting guess:
\[ y_1^{(0)} = y_0 + h \, f(x_0, y_0) \]
Here \(f(x_0, y_0) = 0 + 1 = 1\), so
\[ y_1^{(0)} = 1 + 0.05 \times 1 = 1.05 \]

Step 3: Apply the corrector formula once (first approximation).
The corrector formula improves the guess by averaging the slope at the start and the estimated slope at the end:
\[ y_1^{(k+1)} = y_0 + \frac{h}{2} \Big[ f(x_0, y_0) + f(x_1, y_1^{(k)}) \Big] \]
Using \(y_1^{(0)} = 1.05\):
\[ f(x_1, y_1^{(0)}) = 0.05 + 1.05 = 1.10 \]
\[ y_1^{(1)} = 1 + \frac{0.05}{2} (1 + 1.10) = 1 + 0.025 \times 2.10 = 1.0525 \]
This \(y_1^{(1)} = 1.0525\) is the first approximation from the corrector.

Step 4: Apply the corrector formula a second time (second approximation).
We repeat the same corrector step, now using the improved value \(y_1^{(1)} = 1.0525\):
\[ f(x_1, y_1^{(1)}) = 0.05 + 1.0525 = 1.1025 \]
\[ y_1^{(2)} = 1 + \frac{0.05}{2} (1 + 1.1025) = 1 + 0.025 \times 2.1025 = 1.0525625 \]
This \(y_1^{(2)} = 1.0525625\) is the second approximation.

Final Answer:
Rounding \(1.0525625\) to two decimal places gives \(1.05\). \[ \boxed{1.05} \]