Solve for the function X(s) in the Laplace domain by taking the Laplace transform of the following differential equations with given initial conditions. dx +9x = 161, where x(0) = 0 and (0) = 5

X(s) = 2/[s³(s + 9)]

Step-by-step explanation:

Here is the complete question

Solve for the function X(s) in the Laplace domain by taking the Laplace transform of the following differential equations with given initial conditions.

dx/dt + 9x = 16t²     x(0) = 0 and dx(0)/dt = 5

Solution

dx/dt + 9x = 16t²

Taking the Laplace transform of the differential equation, we have

L{dx/dt + 9x} = L{16t²}

L{dx/dt} + L{9x} = L{16t²}

L{dx/dt} + 9L{x} = 16L{t²}

sL{x} - x(0) + 9L{x}  = 16L{t²}

L{x} = X(s)

L{t²} = 2!/s³

Substituting X(s) and L{t²} into the equation above, we have

sX(s) - x(0) + 9X(s) = 2!/s³

Substituting x(0) = 0 into the equation above, we have

sX(s) - x(0) + 9X(s) = 2!/s³

sX(s) - 0 + 9X(s) = 2!/s³

sX(s) + 9X(s) = 2!/s³

Factorizing X(s), we have

(s + 9)X(s) = 2!/s³

X(s) = 2/[s³(s + 9)]

So X(s) in the Laplace domain is

X(s) = 2/[s³(s + 9)]

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you

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