Find out the answer to "Using Convolution, Solve the Volterra Integral Equation: y(x)=1−e−x+∫0xy(x−t)sin⁡x"

Name: Using Convolution, Solve the Volterra Integral Equation: y(x)=1−e−x+∫0xy(x−t)sin⁡x dt
Uploaded: 2022-02-23T17:22:57+00:00
Description: Using Convolution, Solve the Volterra Integral Equation: y(x)=1−e−x+∫0xy(x−t)sin⁡x dt

melodykap

Answered question

2021-01-17

Using Convolution, Solve the Volterra Integral Equation:

y (x) = 1 - e^{- x} + \int_{0}^{x} y (x - t) \sin x d t

Answer & Explanation

unett

Skilled2021-01-18Added 119 answers

Step 1
Given:

y (x) = 1 - e^{- x} + \int_{0}^{x} y (x - t) \sin x d t

taking Laplace transformation

Y (s) = \frac{1}{s} - \frac{1}{s + 1} + L {\int_{0}^{x} y (x - t) \sin x d t}

Y (s) = \frac{1}{s} - \frac{1}{s + 1} + Y (s) L (\sin x)

Y (s) = \frac{1}{s} - \frac{1}{s + 1} + Y (s) \frac{1}{1 + s^{2}}

Y (s) = (1 - \frac{1}{s^{2} + 1}) = \frac{1}{s (s + 1)}

Y (s) = \frac{s^{2} + 1}{s^{3} (s + 1)} = \frac{(s + 1)^{2} - 2 s}{s^{3} (s + 1)}

Y (s) = \frac{s + 1}{s^{3}} - \frac{2}{s^{2} (s + 1)}

Y (s) = \frac{1}{s^{2}} + \frac{1}{s^{3}} - \frac{2}{s^{2} (s + 1)}

Step 2

Taking Inverse Laplace Transformation

y (x) = L^{- 1} {\frac{1}{s^{2}}} + L^{- 1} {\frac{2}{s^{2 + 1}}} - 2 L^{- 1} {\frac{1}{s^{2} (s + 1)}}

= x + \frac{1}{2} x^{2} + \int_{0}^{x} \int_{0}^{x} (e^{- x} d x) d x

= x + \frac{1}{2} x^{2} + \int_{0}^{x} (e^{x} - 1) d x

= x + \frac{1}{2} x^{2} - {[e^{- x} + x]}_{0}^{x}

= x + \frac{1}{2} x^{2} - e^{- x} - x + 1

y (x) = x + \frac{1}{2} x^{2} - e^{- x}

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