Question

Use theorem 7. 4. 2 to evaluate the given laplace transform. do not evaluate the convolution integral before transforming. (write your answer as a function of s. ) ℒ t et − d 0

Answer 1

With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

So, According to this theorem the equation becomes the

$\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).$

Then after solving, it become and with theorem it says that the

$\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .$

Hence by this way the given equation with convolution theorem is proved.

So, With convolution theorem the equation is proved.

Learn more about convolution theorem here

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