Question

1. (calculator not allowed)

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

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, \(x_{n}\) and \(y_{n}\) are the current values of \(x\) and \(y\), and \(f(x,y)=\frac{dy}{dx}\). Here, \(h = 0.5\), \(x_{0}=1\), \(y_{0}=2\), and \(f(x,y)=x + y\).

Step2: First iteration (\(n = 0\))

Calculate \(y_{1}\):
Substitute \(x_{0}=1\), \(y_{0}=2\) and \(h = 0.5\) into the Euler's formula.
\(y_{1}=y_{0}+h\cdot(x_{0}+y_{0})\)
\(y_{1}=2+0.5\times(1 + 2)\)
\(y_{1}=2+0.5\times3=2 + 1.5=3.5\)
And \(x_{1}=x_{0}+h=1 + 0.5=1.5\)

Step3: Second iteration (\(n = 1\))

Calculate \(y_{2}\):
Substitute \(x_{1}=1.5\), \(y_{1}=3.5\) and \(h = 0.5\) into the Euler's formula.
\(y_{2}=y_{1}+h\cdot(x_{1}+y_{1})\)
\(y_{2}=3.5+0.5\times(1.5 + 3.5)\)
\(y_{2}=3.5+0.5\times5\)
\(y_{2}=3.5 + 2.5\)

Answer:

\(6\)