Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.2 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.2

Question 1.

Express each of the following physical statements in the form of differential equation.

(i) Radium decays at a rate proportional to the amount Q present.

(ii) The population P of a city increases at a rate propotional to the product of population and to the difference between 5,00,000 and the population.

(iii) For a certain substance, the rate of change of vapor pressure P with respect to temperature T is proportional to the vapor pressure and inversely proportional to the square of the temperature.

(iv) A saving amount pays 8% interest per year compound continuously. In addition, the income from another investment is credited to the amount continuously at the rate of Rs 400 per year.

Solution:

(i) If at anytime t, The amount of Radium present is Q. The rate at which Q is decreasing is \(\frac { dQ }{ dt }\) and this is negative. This rate of decrease or decay is found to be proportional to Q itself. Hence we have the law, \(\frac { dQ }{ dt }\) = -kQ where k is a positive constant. Therefore \(\frac { dQ }{ dt }\) + kQ = 0 which is a differential equation.

(ii) The rate of change of population increases with respect to time t, is \(\frac { dP }{ dt }\) & the rate of population is proportional to the product of population is \(\frac { dP }{ dt }\) = kP & the also the difference between 5,00,000 & the population is \(\frac { dP }{ dt }\) = kP(5,00,000 – P) is required differential equation.

(iii) The rate of change of vapor pressure P with respect to time t is \(\frac { dP }{ dt }\) & the rate of increase vapor pressure is P at time T is proportibnal to the vapor pressure and also is inversely proportional to the square of the temperature is \(\frac { dP }{ dt }\) = \(\frac { kP }{ T^2 }\)

since \(\frac { k }{ T^2 }\) = α

\(\frac { dP }{ dt }\) = α P ⇒ \(\frac { dP }{ dt }\) – α P = 0 is a required differential equation.

(iv) Let x be the amount. Amount varies from every year, (ie) Amount varies with respect to time t is \(\frac { dx }{ dt }\) & in addition the income from other source credited Rs.400 continuously for every year.

∴ \(\frac { dx }{ dt }\) = \(\frac { 8 }{ 100 }\) × x + 400

⇒ \(\frac { dx }{ dt }\) = \(\frac { 2x }{ 25 }\) + 400

⇒ ⇒ \(\frac { dx }{ dt }\) – \(\frac { 2 }{ 25 }\)x – 400 = 0 is a required differential equation.

Question 2.

Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

Solution:

The rate of change of the radius of the rain drop evaporates with respect to time t is \(\frac { dx }{ dt }\) = -k i.e The drop is decreasing so put negative sign. Therefore \(\frac { dx }{ dt }\) = -k is a required differential equation.