Question

1. (acorn book) let ( y = f(x) ) be the solution to the differential equation ( \frac{dy}{dx}=arcsin(xy) ) with the initial condition ( f(0)=2 ). what is the approximation for ( f(1) ) if eulers method is used, starting at ( x = 0 ) with a step size of ( 0.5 )?

(a) 2 (b) ( 2+\frac{pi}{6} ) (c) ( 2+\frac{pi}{4} ) (d) ( 2+\frac{pi}{2} ) (e) 3

Explanation:

Step1: Recall Euler's Method formula

Euler's Method formula is \(y_{n + 1}=y_{n}+h\cdot f(x_{n},y_{n})\), where \(h\) is the step - size. Here, \(h = 0.5\), \(x_{0}=0\), \(y_{0}=2\), and \(\frac{dy}{dx}=\arcsin(xy)\).

Step2: Calculate \(y_1\)

First, find \(y_1\) when \(n = 0\).
We know that \(x_1=x_0 + h=0 + 0.5=0.5\).
Substitute \(x_0 = 0\) and \(y_0 = 2\) into \(\frac{dy}{dx}=\arcsin(xy)\): \(\frac{dy}{dx}\big|_{x = 0,y = 2}=\arcsin(0\times2)=0\).
Using Euler's formula \(y_1=y_0+h\cdot\frac{dy}{dx}\big|_{x_0,y_0}\), so \(y_1=2+0.5\times0 = 2\).

Step3: Calculate \(y_2\)

Now, \(x_2=x_1+h=0.5 + 0.5 = 1\).
Substitute \(x_1 = 0.5\) and \(y_1 = 2\) into \(\frac{dy}{dx}=\arcsin(xy)\): \(\frac{dy}{dx}\big|_{x = 0.5,y = 2}=\arcsin(0.5\times2)=\arcsin(1)=\frac{\pi}{2}\).
Using Euler's formula \(y_2=y_1+h\cdot\frac{dy}{dx}\big|_{x_1,y_1}\), so \(y_2=2+0.5\times\frac{\pi}{2}=2+\frac{\pi}{4}\).

Answer:

C. \(2+\frac{\pi}{4}\)