Using the Laplace transform, the value o y' − 2y = (t − 4), y(0) = 0 is⇒y(t) = 0 e^-t + u(t -1)e^1-t
Laplace rework is an critical rework approach that is in particular useful in fixing linear normal equations. It unearths very huge applications in regions of physics, electrical engineering, control optics, arithmetic and sign processing.
y' − 2y = (t − 4),
y(0) = 0
Taking the Laplace transformation of the differential equation
⇒sY(s) - y (0) + Y(s) = e-s
⇒(s + 1)Y(s) = (0+ e^-s)/s + 1
⇒y(t) = L^-1{0/s+1} + {e ^-s/s + 1}
⇒y(t) = 0 e^-t + u(t -1)e^1-t
The Laplace remodel method, the feature within the time area is transformed to a Laplace characteristic within the frequency domain. This Laplace feature will be inside the shape of an algebraic equation and it can be solved easily.
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I’m not sure if this is right I hope it helps but you weren’t five to the right with the negative one to the four so you have to go five to the left making it -6 for the first coordinate and then he went 13 to the left for the second coordinate so you have to go 13 to the right 6+13 is 19. So your answer is -6, 19 hope this helps!
Answer:
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Step-by-step explanation:
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It is 2.69333333, to the nearest centimetre the answer is B) 3 cm per second
