Question

Extra Credit Use convolutions to find an explicit function \( y(t) \) such that \[ y(t)=1+\int_{0}^{t} y(x) \sin (t-x) d x . \]

Answer 1

The explicit function
`\( y(t) \)` satisfying the given integral equation is
`\( y(t) = \sin(t) + \cos(t) \)`.

Explanation:

To find
`\( y(t) \)`, we first observe that the integrand involves
`\( y(x) \sin(t-x) \)`, which suggests the use of convolution. The convolution of two functions
`\( f(t) \) and \( g(t) \) is given by \((f * g)(t) = \int_(-\infty)^(\infty) f(x)g(t-x)dx\).`

Applying convolution to the given integral equation, we have
`\(y(t) = 1 + \int_(0)^(t) y(x) \sin(t-x)dx = 1 + (\sin * y)(t)\)`, where
`\(\sin(t)\)` is the kernel function. We need to solve this convolution integral equation.

The convolution of
`\( \sin(t) \) with \( y(t) \)` can be found using the convolution theorem, which states that the Fourier transform of the convolution of two functions is the product of their individual Fourier transforms. The Fourier transform of
`\( \sin(t) \) is \( \pi (\delta(\omega - 1) - \delta(\omega + 1)) \), where \( \delta \)` is the Dirac delta function. The Fourier transform of
`\( y(t) \) is denoted by \( Y(\omega) \)`.

Solving for

`\( Y(\omega) \), we find that \( Y(\omega) = (\pi)/(2) \left((1)/(\omega - 1) - (1)/(\omega + 1)\right) \). Taking the inverse Fourier transform of \( Y(\omega) \) gives \( y(t) = \sin(t) + \cos(t) \)` satisfying the given integral equation.