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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Answer:
(x-5)*(x+5)
Step-by-step explanation:
= x^2 + 5x - 25 - 5x
= x^2 +5x - 5x -25
because +5x - 5x = 0 so
= x^2 + 0 - 25
= x^2 - 25
25 can be written as 5^2
= x^2 - 5^2
we know that a^2 - b^2 = (a-b) * (a+b)
so x^2 - 5^2
= (x-5) * (x+5)
Hope this help you :3
C. Is most likely the better answer
you can do that repeated addition
