Answer:
[tex]y_1=\boxed{-3}[/tex]
[tex]y_2=\boxed{-8}[/tex]
[tex]y_3=\boxed{-16}[/tex]
[tex]y_4=\boxed{-28.5}[/tex]
Step-by-step explanation:
Given:
- [tex]y'=y-2x, \;\;y(3)=0[/tex]
Use Euler’s method with h = 0.5 to find approximate values for the solution of the initial value problem:
[tex]y'=y-2x, \;\;y(3)=0[/tex]
at x = 3.5, 4, 4.5, 5.
Therefore:
[tex]f(x,y)=y-2x, \quad x_0=3 \;\;\text{and}\; \;y_0=0[/tex]
Euler's Method
[tex]\boxed{y_{n+1}=y_n+hf(x_n,y_n)}[/tex]
[tex]\begin{aligned}\implies y_1&=y_0+hf(x_0,y_0)\\&=0+(0.5)f(3,0)\\&=0+0.5[0-2(3)]\\&=0+0.5(-6)\\&=-3\end{aligned}[/tex]
[tex]\begin{aligned}\implies y_2&=y_1+hf(x_1,y_1)\\&=-3+(0.5)f(3.5,-3)\\&=-3+0.5[-3-2(3.5)]\\&=-3+0.5(-10)\\&=-3-5\\&=-8\end{aligned}[/tex]
[tex]\begin{aligned}\implies y_3&=y_2+hf(x_2,y_2)\\&=-8+(0.5)f(4,-8)\\&=-8+0.5[-8-2(4)]\\&=-8+0.5(-16)\\&=-8-8\\&=-16\end{aligned}[/tex]
[tex]\begin{aligned}\implies y_4&=y_3+hf(x_3,y_3)\\&=-16+(0.5)f(4.5,-16)\\&=-16+0.5[-16-2(4.5)]\\&=-16+0.5(-25)\\&=-16-12.5\\&=-28.5\end{aligned}[/tex]
[tex]\begin{array}{|c|c|c|}\cline{1-3} \text{Step}\;(n) & x_n&y_n\\\cline{1-3} 0&3&0\\\cline{1-3} 1&3.5&-3\\\cline{1-3} 2&4&-8\\\cline{1-3} 3&4.5&-16\\\cline{1-3} 4&5&-28.5\\\cline{1-3} \end{array}[/tex]
