Solve the following differential equations using the substitutions given.

(a)

(b)

A rectangular tank with its base horizontal is filled with water to a depth *h *at time *t* = 0. Water leaks out of the tank from a small hole in the base at a rate proportional to the square root of the depth of the water. If the depth of water is ½*h* at time* T,* find the further time, in terms of *T*, it will take before the tank is empty.

A researcher is investigating the spread of a certain disease in a town with a population of 3000 people. The researcher suggests that *I*, the number of people infected by the disease at time *t *days satisfies the differential equation , where *k* is a positive constant.

**(i) **Given that *I *= 30 when *t *= 0, show that .

**(ii) **It is further observed that *I *= 240 when *t *= 7 find the time it takes for 90% of the population to be infected by the disease.

**(iii) **State, in the context of this question, one assumption needed to model the spread of the disease in the town by the given differential equation.

(iii) Possible Answer:

Assume that the total population of the town is 3000 during the spread of the disease.

Or: Assume that a person infected by the disease will remain infected by the disease.

Or: Assume that everyone in the town has an equal chance of being infected by the disease.

(a)Without using a calculator, solve the inequality

(b) Deduce the range of values of x that satisfies.

Solutions: (a): x≤-6 or 1≤x<3 or x>3

(b): 0<x≤e^{-6} or e≤x<e^{3} or x>e^{3}

(a)

(b)

lnx ≤ -6

x ≤ e^{-6}

1≤ lnx <3

e≤x<e^{3}

lnx>3

x>e^{3}

The figure shows the curve x = y^{2}-9. Find the area of the region bounded by the curve, y-axis and the line y=4

Find the total area enclosed by the curve y = x(x+3)(2-x), the x-axis and the line x=4

solution: 21(1/12) units

x(2x – x^{2} + 6 – 3x)

=6x – x^{3} – x^{2}

Differentiate y=5x(e^{3x}) with respect to x and hence find ∫x(e^{3x}) dx

Solution: 1/3(x)(e^{3x}) – 1/9(e^{3x}) + C

Y = 5x e^{3x}

dy/dx = 5x(e^{3x})(3) + 5 e^{3x}

dy/dx = 15x(e^{3x}) + 5(e^{3x})

∫15x(e^{3x}) + 5(e^{3x}) dx = 5x(e^{3x})

∫15x(e^{3x}) dx + (5(e^{3x}))/3 = 5x(e^{3x})

15∫x(e^{3x}) dx = 5x(e^{3x}) – 5/3(e^{3x})

∫x(e^{3x}) dx = 1/3(x)(e^{3x}) – 1/9(e^{3x}) + C