2 years ago

PLEASE HELP. WILL GIVE BRAINLIEST.

Mathematics

1 answer:

mezya [45]2 years ago

8 0

Answer:

Step-by-step explanation:

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3 years ago

Find the inverse Laplace transform f(t) of the function F(s). Write uc for the Heaviside function that turns on at c, not uc(t).

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Answer:

F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

Step-by-step explanation:

We have given $F(S)=\frac{7e^{-7s}}{s^2-49}$

Now $F(S)=e^{-7s}G(s)$

Here $G(S)=\frac{7}{S^2-49}$

Now first find the Laplace inverse of G(S)

Using partial fraction

$\frac{7}{(s+7)(s-7)}=\frac{A}{(S+7)}+\frac{B}{S-7}$

$7=A(S-7)+B(S+7)$

On comparing the coefficient

$A=\frac{1}{2}$ and $B=\frac{-1}{2}$

On putting the value of A and B

$G(S)=\frac{-1}{2(S+7)}+\frac{1}{2(S+7)}$

Taking inverse Laplace

$G(t)=\frac{-1}{2}e^{7t}+\frac{1}{2}e^{-7t}$

Now in G(s) there is onether term $e^{-7s}$

So F(t) $=\frac{-1}{2}e^{7(t-7)}+\frac{1}{2}e^{-7(t-7)}$

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What value of x makes this equation true?

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Step-by-step explanation:

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