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)]

i need points

step-by-step explanation:

sorry bubs

answer:

total cost for tiles and paints is $924.  

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

given: 1m squared floor tiles costs $3. so,

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

therefore, the combined total cost of tiles and paint is $924.