91Ó°ÊÓ

The use of Laplace transformation is to convertdifferential equationsinto algebraic equations. The formula for Laplace transform is

Where, F(s) = Laplace Transform

S = complex number

t = real number >=0

t’ = first derivative of the function f(t)

Since the solution removes 6L/min we have that the solution increases by 12-6=6L, from where it follows that the solution can be represented as linear function 500+6t.

Therefore, the output rate is given as

\(\begin{array}{c}{\rm{\;or\;}} = \frac{m}{{500 + 6t}} \cdot 6\\ = \frac{{3m}}{{250 + 3t}}\end{array}\)

Where \(m(t)\) be the mass of salt in the tank at time t.

Input rate is the same as in the problem 33

\({\rm{\;ir\;}} = 0.48 + 0.72u(t - 10)\)

Now, we get the differential equation:

\(\begin{array}{c}m' = {\rm{\;ir\;}} - {\rm{\;or\;}}\\m' + \frac{{3m}}{{250 + 3t}} = 0.48 + 0.72u(t - 10)\\(250 + 3t)m' + 3m = (250 + 3t)(0.48 + 0.72u(t - 10))\end{array}\)

which is the differential equation with polynomial coefficients and it can be solved by Laplace transform.

Hence, the answer is:

Yes, Laplace transforms can be used to solve the problem.