For the given function f(t) = (2t + 1) using definition of Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].
As given in the question,
Given function is equal to :
f(t) = 2t + 1
Simplify the given function using definition of Laplace transform we have,
L(f(t))s = 
= ![\int\limits^\infty_0[2t +1] e^{-st} dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%5Cinfty_0%5B2t%20%2B1%5D%20e%5E%7B-st%7D%20dt)
= 
= 2 L(t) + L(1)
L(1) = 
= (-1/s) ( 0 -1 )
= 1/s , ( s > 0)
2L ( t ) = 
= ![2[t\int\limits^\infty_0 e^{-st} - \int\limits^\infty_0 ({(d/dt)(t) \int\limits^\infty_0e^{-st} \, dt )dt]](https://tex.z-dn.net/?f=2%5Bt%5Cint%5Climits%5E%5Cinfty_0%20e%5E%7B-st%7D%20-%20%5Cint%5Climits%5E%5Cinfty_0%20%28%7B%28d%2Fdt%29%28t%29%20%5Cint%5Climits%5E%5Cinfty_0e%5E%7B-st%7D%20%5C%2C%20dt%20%29dt%5D)
= 2/ s²
Now ,
L(f(t))s = 2 L(t) + L(1)
= 2/ s² + 1/s
Therefore, the solution of the given function using Laplace transform the required solution is L(f(t))s = [ ( 2/s²) + ( 1/s) ].
Learn more about Laplace transform here
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The first step is to write this equation into general form. The
general form of an equation is:
ax^2 + bx + c = 0
To make this equation to general form, you have to simplify
the equation first.
2/3(x-4) (x+5) = 1
2/3 (x^2 + 5x – 4x – 20) = 1
2/3(x^2 + x -20) = 1
2/3x^2 + 2/3x – 40/3 = 1
2/3x^2 + 2/3x – 40/3 – 1 = 0
2/3x^2 +2/3x – 43/3 = 0
Therefore, a = 2/3 ; b = 2/3 ; c = -43/3
Answer:
Can you add a picture to your question?
Step-by-step explanation:
Step-by-step explanation:
<h2>
<em><u>The following graph shows the ... ... or not to continue the recent lunch special promotion at his restaurant. ... earned, y, from the number of lunch specials ordered per hour, x.</u></em></h2>
Answer:
y=3x-1
Step-by-step explanation:
y=mx+b
y=3x+b
y=3x-1
