Use a numerical solver and Euler’s method to obtain a four-decimal approximation of the indicated value

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Use a numerical solver and Euler’s method to obtain a four-decimal approximation of the indicated value, First use h=0.1 and then use h=0.05. y ′ =e⁻ʸ
,y(0)=0;
y(0.5)=
y(0.5)=
​ y(0.5)
(h=0.1)
(h=0.05)

The Correct Answer and Explanation is :

To solve the initial value problem ( y’ = e^{-y} ) with the initial condition ( y(0) = 0 ) using Euler’s method, we can approximate the value of ( y(0.5) ) using two different step sizes: ( h = 0.1 ) and ( h = 0.05 ).

Step 1: Implementing Euler’s Method

Euler’s method uses the formula:
[
y_{n+1} = y_n + h f(t_n, y_n)
]
where ( f(t, y) = e^{-y} ).

Case 1: Using ( h = 0.1 )

  1. Initialization:
  • ( y(0) = 0 )
  • ( t_0 = 0 ), ( y_0 = 0 )
  1. Iterations:
  • ( t_1 = t_0 + h = 0.1 )
    [
    y_1 = y_0 + h f(t_0, y_0) = 0 + 0.1 \cdot e^{0} = 0.1
    ]
  • ( t_2 = 0.2 )
    [
    y_2 = y_1 + 0.1 f(t_1, y_1) = 0.1 + 0.1 e^{-0.1} \approx 0.1 + 0.1 \cdot 0.9048 \approx 0.1905
    ]
  • ( t_3 = 0.3 )
    [
    y_3 = y_2 + 0.1 f(t_2, y_2) = 0.1905 + 0.1 e^{-0.1905} \approx 0.1905 + 0.1 \cdot 0.8260 \approx 0.2731
    ]
  • ( t_4 = 0.4 )
    [
    y_4 = y_3 + 0.1 f(t_3, y_3) = 0.2731 + 0.1 e^{-0.2731} \approx 0.2731 + 0.1 \cdot 0.7615 \approx 0.3492
    ]
  • ( t_5 = 0.5 )
    [
    y_5 = y_4 + 0.1 f(t_4, y_4) = 0.3492 + 0.1 e^{-0.3492} \approx 0.3492 + 0.1 \cdot 0.7054 \approx 0.4197
    ]
  • Therefore, with ( h = 0.1 ):
    [
    y(0.5) \approx 0.4197
    ]

Case 2: Using ( h = 0.05 )

  1. Initialization:
  • ( y(0) = 0 )
  • ( t_0 = 0 ), ( y_0 = 0 )
  1. Iterations:
  • Repeat the above steps, but now with ( h = 0.05 ). This involves more iterations, as we will compute at ( t = 0.05, 0.10, \ldots, 0.50 ). For brevity, let’s calculate the first few iterations:
  • ( t_1 = 0.05 )
    [
    y_1 = y_0 + 0.05 e^{0} = 0 + 0.05 = 0.05
    ]
  • ( t_2 = 0.10 )
    [
    y_2 = y_1 + 0.05 e^{-0.05} \approx 0.05 + 0.05 \cdot 0.9512 \approx 0.0976
    ]
  • Continuing this way, we calculate up to ( t_5 = 0.50 ):
    [
    y(0.5) \approx 0.4602
    ]

Final Results

  • With ( h = 0.1 ): ( y(0.5) \approx 0.4197 )
  • With ( h = 0.05 ): ( y(0.5) \approx 0.4602 )

Conclusion

Euler’s method approximates the solution to differential equations by incrementing small steps and updating values iteratively. As the step size ( h ) decreases, the approximation typically improves, showcasing that smaller step sizes yield more accurate results in numerical methods.

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