Question

Use euler’s method. i. e. to find the approximate values of the solution for yʹ=y(3−ty), y(0)=0. 5,h=0. 1, 0≤t≤0. 5

Answer 1

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,

$y_{n+1}=y_n+hf_n$

where $f_n=f(t_n,y_n)$

For, t = 0, y = 0,

$y_{1}=y_0+hf_0$ and

$f_0=f(t_0,y_0)$

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, $y_{1}=y_0+hf_0$ becomes,

⇒ y₁ = 0.5 + (0.1)(1.5)

⇒ y₁ = 0.5 + 0.15

⇒ y₁ = 0.65

For, t = 1, y = 1,

$y_{2}=y_1+hf_1$ and

$f_1=f(t_1,y_1)$

⇒ 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, $y_{2}=y_1+hf_1$ becomes,

⇒ y₂ = 0.65+(0.1)(1.90)

⇒ y₂ = 0.84

For t = 2, y = 2,

$y_{3}=y_2+hf_2$ and

$f_2=f(t_2,y_2)$

⇒ 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, $y_{3}=y_2+hf_2$

⇒ y₃ = 0.84+(0.1)(2.3788)

⇒ y₃ = 1.0778

For t = 3, y = 3,

$y_{4}=y_3+hf_3$ and

$f_3=f(t_3,y_3)$

⇒ 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, $y_{4}=y_3+hf_3$ becomes

⇒ y₄ = 1.0778+(0.1)(2.8848)

⇒ y₄ = 1.3584

For t = 4, y = 4,

$y_{5}=y_4+hf_4$ and

$f_4=f(t_4,y_4)$

⇒ 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, $y_{5}=y_4+hf_4$ becomes

⇒ y₅ = 1.3584 + (0.1)(3.3372)

⇒ y₅ = 1.6921

For t = 5, y = 5,

$y_{6}=y_5+hf_5$ and

$f_5=f(t_5,y_5)$

⇒ 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, $y_{6}=y_5+hf_5$ 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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