Question

Use Euler’s method to solve the I.V.P.
y'=(x-y)/2, on [0,3], with y(0) = 1, and h = 0.5

Answer 1

Given the initial value problem, $y'=\dfrac{x-y}{2}$ with initial condition $y(0)=1$.

The Euler's method states that,

$y_{n+1} = y_{n}+h\cdot f(x_{n},y_{n})$, where $x_{n+1} = x_{n}+h$.

Here, $h = 0.5, x_{0} = 0, y_{0} = 1, f(x,y) = y' = \dfrac{x - y}{2}$ .

Step 1:

$x_{1} = x_{0} + h = 0.5\\ y_{1}=y(0.5)=y_{0}+h\cdot f(x_{0},y_{0})\\ = 1+ 0.5 \left(\dfrac{0-1}{2}\right) = 0.75$

Step 2:

$x_{2} = x_{1} + h = 1\\ y_{2}=y(1)=y_{1}+h\cdot f(x_{1},y_{1})\\ = 0.75+ 0.5 \left(\dfrac{0.5-0.75}{2}\right) = 0.6875$

Step 3:

$x_{3} = x_{2} + h = 1.5\\ y_{3}=y(1.5)=y_{2}+h\cdot f(x_{2},y_{2})\\ = 0.6875 + 0.5 \left(\dfrac{1-0.6875}{2}\right) = 0.765625$

Step 4:

$x_{4} = x_{3} + h = 2\\ y_{4}=y(2)=y_{3}+h\cdot f(x_{3},y_{3})\\ = 0.765625 + 0.5 \left(\dfrac{1.5-0.765625}{2}\right) = 0.94921875$

Step 5:

$x_{5} = x_{4} + h = 2.5\\ y_{5}=y(2.5)=y_{4}+h\cdot f(x_{4},y_{4})\\ = 0.94921875 + 0.5 \left(\dfrac{2-0.94921875}{2}\right)\\ = 1.2119140625$

Step 6:

$x_{6} = x_{5} + h = 3\\ y_{6}=y(3)=y_{5}+h\cdot f(x_{5},y_{5})\\ = 1.2119140625 + 0.5 \left(\dfrac{2.5-1.2119140625}{2}\right)\\ = 1.533935546875$