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The approximate values of the solution by Euler's method is y₁=0.65, y₂=0.84, y₃=1.0778, y₄=1.3584, y₅=1.6921, y₆=2.05657.
In this question,
The differential equation is
yʹ=y(3−ty) ------- (1)
Here, y(0)=0. 5,h=0. 1, 0≤t≤0. 5.
By Euler's method,
[tex]y_{n+1}=y_n+hf_n[/tex]
where [tex]f_n=f(t_n,y_n)[/tex]
For, t = 0, y = 0,
[tex]y_{1}=y_0+hf_0[/tex] and
[tex]f_0=f(t_0,y_0)[/tex]
Substitute in equation 1,
⇒ f(0,0.5) = 0.5(3-(0)(0.5))
⇒ f(0,0.5) = 0.5(3-0)
⇒ f(0,0.5) = 0.5(3)
⇒ f(0,0.5) = 1.5
Then, [tex]y_{1}=y_0+hf_0[/tex] becomes,
⇒ y₁ = 0.5 + (0.1)(1.5)
⇒ y₁ = 0.5 + 0.15
⇒ y₁ = 0.65
For, t = 1, y = 1,
[tex]y_{2}=y_1+hf_1[/tex] and
[tex]f_1=f(t_1,y_1)[/tex]
⇒ f(0.1,0.65) = 0.65(3-(0.1)(0.65))
⇒ f(0.1,0.65) = 0.65(3-0.065)
⇒ f(0.1,0.65) = 1.90
Then, [tex]y_{2}=y_1+hf_1[/tex] becomes,
⇒ y₂ = 0.65+(0.1)(1.90)
⇒ y₂ = 0.84
For t = 2, y = 2,
[tex]y_{3}=y_2+hf_2[/tex] and
[tex]f_2=f(t_2,y_2)[/tex]
⇒ f(0.2,0.84) = 0.84(3-(0.2)(0.84))
⇒ f(0.2,0.84) = 0.84(3-0.168)
⇒ f(0.2,0.84) = 2.3788
Then, [tex]y_{3}=y_2+hf_2[/tex]
⇒ y₃ = 0.84+(0.1)(2.3788)
⇒ y₃ = 1.0778
For t = 3, y = 3,
[tex]y_{4}=y_3+hf_3[/tex] and
[tex]f_3=f(t_3,y_3)[/tex]
⇒ f(0.3,1.0778) = 1.0778(3-(0.3)(1.0778))
⇒ f(0.3,1.0778) = 1.0778(3-0.32334)
⇒ f(0.3,1.0778) = 2.8848
Then, [tex]y_{4}=y_3+hf_3[/tex] becomes
⇒ y₄ = 1.0778+(0.1)(2.8848)
⇒ y₄ = 1.3584
For t = 4, y = 4,
[tex]y_{5}=y_4+hf_4[/tex] and
[tex]f_4=f(t_4,y_4)[/tex]
⇒ f(0.4,1.3584) = 1.3584(3-(0.4)(1.3584))
⇒ f(0.4,1.3584) = 1.3584(3-0.5433)
⇒ f(0.4,1.3584) = 3.3372
Then, [tex]y_{5}=y_4+hf_4[/tex] becomes
⇒ y₅ = 1.3584 + (0.1)(3.3372)
⇒ y₅ = 1.6921
For t = 5, y = 5,
[tex]y_{6}=y_5+hf_5[/tex] and
[tex]f_5=f(t_5,y_5)[/tex]
⇒ f(0.5,1.6921) = 1.6921(3-(0.5)(1.6921))
⇒ f(0.5,1.6921) = 1.6921(3-0.84605)
⇒ f(0.5,1.6921) = 3.6447
Then, [tex]y_{6}=y_5+hf_5[/tex] becomes
⇒ y₆ = 1.6921 + (0.1)(3.6447)
⇒ y₆ = 2.05657
Hence we can conclude that the approximate values of the solution by Euler's method is y₁=0.65, y₂=0.84, y₃=1.0778, y₄=1.3584, y₅=1.6921, y₆=2.05657.
Learn more about Euler's method here
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