Maharashtra Board Class 12 Maths Part 1 Chapter 8 Differential Equation 8.6 Solutions

Question 1. In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Answer: Solution: Let x be the number of bacteria in the culture at time t. Then the rate of increase is \( \frac{dx}{dt} \) which is proportional to x.
\( \implies \frac{dx}{dt} \propto x \)
\( \implies \frac{dx}{dt} = kx \), where k is a constant
\( \implies \frac{dx}{x} = k dt \) On integrating, we get
\( \int \frac{dx}{x} = \int k dt \)
\( \implies \log x = kt + c \) Initially, i.e. when t = 0, let x = x₀
\( \implies \log x_0 = k \times 0 + c \)
\( \implies c = \log x_0 \)
\( \implies \log x = kt + \log x_0 \)
\( \implies \log x - \log x_0 = kt \)
\( \implies \log\left(\frac{x}{x_0}\right) = kt \ldots \ldots (1) \) Since the number doubles in 4 hours, i.e. when t = 4, x = 2x₀
\( \implies \log\left(\frac{2x_0}{x_0}\right) = 4k \)
\( \implies \log 2 = 4k \)
\( \implies k = \frac{1}{4} \log 2 \)
\( \therefore \) (1) becomes, \( \log\left(\frac{x}{x_0}\right) = \frac{t}{4} \log 2 \) When t = 12, we get
\( \log\left(\frac{x}{x_0}\right) = \frac{12}{4} \log 2 = 3 \log 2 \)
\( \implies \log\left(\frac{x}{x_0}\right) = \log 2^3 \)
\( \implies \log\left(\frac{x}{x_0}\right) = \log 8 \)
\( \implies \frac{x}{x_0} = 8 \)
\( \implies x = 8x_0 \)
\( \therefore \) the number of bacteria will be 8 times the original number in 12 hours.
In simple words: The bacterial growth follows an exponential model. By using the given information that the number doubles in 4 hours, we find the constant of proportionality. Substituting t=12 hours into the derived equation, we calculate that the bacteria will increase 8 times their initial number.

🎯 Exam Tip: Remember to clearly define variables and state the proportionality relation. Ensure all logarithmic and exponential properties are applied correctly in calculations. Pay attention to the initial conditions and subsequent time points.

Question 2. If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? \( \left(\text{Given:} \sqrt{\frac{3}{2}} = 1.2247\right) \)
Answer: Solution: Let P be the population of the city at time t. Then \( \frac{dP}{dt} \), the rate of increase of population, is proportional to P.
\( \implies \frac{dP}{dt} \propto P \)
\( \implies \frac{dP}{dt} = kP \), k is a constant
\( \implies \frac{dP}{P} = k dt \) Integrating, we get
\( \int \frac{dP}{P} = \int k dt \)
\( \implies \log P = kt + c \) Initially, i.e. when t = 0, P = 40000
\( \implies \log 40000 = 0 + c \)
\( \implies c = \log 40000 \)
\( \implies \log P = kt + \log 40000 \)
\( \implies \log P - \log 40000 = kt \)
\( \implies \log\left(\frac{P}{40000}\right) = kt \ldots \ldots (1) \) When t = 40, P = 60000
\( \implies \log\left(\frac{60000}{40000}\right) = 40k \)
\( \implies \log\left(\frac{3}{2}\right) = 40k \)
\( \implies k = \frac{1}{40} \log\left(\frac{3}{2}\right) \)
\( \therefore \) (1) becomes, \( \log\left(\frac{P}{40000}\right) = \frac{t}{40} \log\left(\frac{3}{2}\right) \)
\( \implies \log\left(\frac{P}{40000}\right) = \log\left(\left(\frac{3}{2}\right)^{\frac{t}{40}}\right) \)
\( \implies \frac{P}{40000} = \left(\frac{3}{2}\right)^{\frac{t}{40}} \) We have to find P in another 20 years i.e. at t = 40 + 20 = 60 If t = 60, then
\( \frac{P}{40000} = \left(\frac{3}{2}\right)^{\frac{60}{40}} \)
\( \implies \frac{P}{40000} = \left(\frac{3}{2}\right)^{\frac{3}{2}} \)
\( \implies \frac{P}{40000} = \frac{3}{2} \sqrt{\frac{3}{2}} \)
\( \implies P = 40000 \times \frac{3}{2} \times 1.2247 \) [By data]
\( \implies P = 20000 \times 3 \times 1.2247 \)
\( \implies P = 60000 \times 1.2247 \)
\( \implies P = 73482 \)
\( \therefore \) population after 60 years will be 73482.
In simple words: The population growth follows an exponential model. We first determine the growth constant 'k' using the initial population and its increase over 40 years. Then, we use this constant to predict the population after an additional 20 years (total 60 years), utilizing the provided value for the square root.

🎯 Exam Tip: For population growth problems, remember to set up the differential equation correctly and integrate it. Pay close attention to the time intervals and initial conditions when solving for the constant and predicting future values.

Question 3. The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after \( \frac{5}{2} \) hours. \( (\text{Given: } \sqrt{2} = 1.414) \)
Answer: Solution: Let x be the number of bacteria at time t. Then the rate of increase is \( \frac{dx}{dt} \) which is proportional to x.
\( \implies \frac{dx}{dt} \propto x \)
\( \implies \frac{dx}{dt} = kx \), where k is a constant
\( \implies \frac{dx}{x} = k dt \) On integrating, we get
\( \int \frac{dx}{x} = \int k dt \)
\( \implies \log x = kt + c \) Initially, i.e. when t = 0, x = 1000
\( \implies \log 1000 = k \times 0 + c \)
\( \implies c = \log 1000 \)
\( \implies \log x = kt + \log 1000 \)
\( \implies \log x - \log 1000 = kt \)
\( \implies \log\left(\frac{x}{1000}\right) = kt \ldots \ldots (1) \) Now, when t = 1, x = 2 \( \times \) 1000 = 2000
\( \therefore \log\left(\frac{2000}{1000}\right) = k \times 1 \)
\( \implies \log 2 = k \)
\( \implies k = \log 2 \)
\( \therefore \) (1) becomes, \( \log\left(\frac{x}{1000}\right) = t \log 2 \) If t = \( \frac{5}{2} \), then
\( \log\left(\frac{x}{1000}\right) = \frac{5}{2} \log 2 \)
\( \implies \log\left(\frac{x}{1000}\right) = \log\left(2^{\frac{5}{2}}\right) \)
\( \implies \frac{x}{1000} = 2^{\frac{5}{2}} \)
\( \implies \frac{x}{1000} = 2^2 \sqrt{2} \)
\( \implies \frac{x}{1000} = 4\sqrt{2} \)
\( \implies \frac{x}{1000} = 4 \times 1.414 \)
\( \implies \frac{x}{1000} = 5.656 \)
\( \implies x = 5.656 \times 1000 \)
\( \implies x = 5656 \)
\( \therefore \) number of bacteria after \( \frac{5}{2} \) hours = 5656.
In simple words: This problem involves exponential growth of bacteria. We first establish the constant of proportionality 'k' using the initial count and the doubling time. Then, we substitute the new time (2.5 hours) into the derived exponential growth formula to calculate the final number of bacteria.

🎯 Exam Tip: Ensure that you correctly interpret 'doubles in 1 hour' to find the growth constant. Fractional exponents like \( 2^{5/2} \) should be simplified carefully, often by separating into an integer power and a square root, for accurate calculation.

Question 4. Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Answer: Solution: Let P be the population of the city at time t. Then \( \frac{dP}{dt} \), the rate of increase of population, is proportional to P.
\( \implies \frac{dP}{dt} \propto P \)
\( \implies \frac{dP}{dt} = kP \), where k is a constant.
\( \implies \frac{dP}{P} = k dt \) On integrating, we get
\( \int \frac{dP}{P} = \int k dt \)
\( \implies \log P = kt + c \) Initially, i.e. when t = 0, P = 30000
\( \implies \log 30000 = k \times 0 + c \)
\( \implies c = \log 30000 \)
\( \implies \log P = kt + \log 30000 \)
\( \implies \log P - \log 30000 = kt \)
\( \implies \log\left(\frac{P}{30000}\right) = kt \ldots \ldots (1) \) Now, when t = 40, P = 40000
\( \implies \log\left(\frac{40000}{30000}\right) = k \times 40 \)
\( \implies \log\left(\frac{4}{3}\right) = 40k \)
\( \implies k = \frac{1}{40} \log\left(\frac{4}{3}\right) \)
\( \therefore \) (1) becomes, \( \log\left(\frac{P}{30000}\right) = \frac{t}{40} \log\left(\frac{4}{3}\right) \)
\( \implies \log\left(\frac{P}{30000}\right) = \log\left(\left(\frac{4}{3}\right)^{\frac{t}{40}}\right) \)
\( \implies \frac{P}{30000} = \left(\frac{4}{3}\right)^{\frac{t}{40}} \)
\( \implies P = 30000 \left(\frac{4}{3}\right)^{\frac{t}{40}} \)
\( \therefore \) the population of the city at time t is \( 30000 \left(\frac{4}{3}\right)^{\frac{t}{40}} \).
In simple words: We model the city's population growth as proportional to its current size, leading to an exponential function. Using the given population increase over 40 years, we determine the growth constant. Finally, we express the population 'P' as a function of time 't'.

🎯 Exam Tip: When deriving the population function, ensure that the constant 'k' is accurately determined from the given data. The final answer should be an expression for P in terms of t, clearly showing the initial population and the derived growth factor.

Question 5. The rate of depreciation \( \frac{dV}{dt} \) of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was Rs. 8,00,000 and its value decreased Rs. 1,00,000 in the first year. Find the value after 6 years.
Answer: Solution: Let V be the value of the machine at the end of t years. Then \( \frac{dV}{dt} \), the rate of depreciation, is inversely proportional to \( (t + 1)^2 \).
\( \implies \frac{dV}{dt} \propto \frac{1}{(t+1)^2} \) Since it's depreciation, the rate of change is negative.
\( \implies \frac{dV}{dt} = \frac{-k}{(t+1)^2} \), k > 0 is a constant
\( \implies dV = \frac{-k}{(t+1)^2} dt \) On integrating, we get
\( \int dV = -k \int \frac{dt}{(t+1)^2} \)
\( \implies V = -k \left[\frac{-1}{t+1}\right] + c \)
\( \implies V = \frac{k}{t+1} + c \) Initially, i.e. when t = 0, V = 800000
\( \implies 800000 = \frac{k}{0+1} + c = k + c \ldots \ldots (1) \) Now, when t = 1, V = 800000 - 100000 = 700000
\( \implies 700000 = \frac{k}{1+1} + c = \frac{k}{2} + c \ldots \ldots (2) \) Subtracting (2) from (1), we get
\( 800000 - 700000 = (k + c) - \left(\frac{k}{2} + c\right) \)
\( \implies 100000 = k - \frac{k}{2} \)
\( \implies 100000 = \frac{k}{2} \)
\( \implies k = 200000 \) Substitute k in (1):
\( \implies 800000 = 200000 + c \)
\( \implies c = 800000 - 200000 \)
\( \implies c = 600000 \) So, the equation for V is:
\( V = \frac{200000}{t+1} + 600000 \) When t = 6,
\( V = \frac{200000}{6+1} + 600000 \)
\( \implies V = \frac{200000}{7} + 600000 \)
\( \implies V = 28571.4285... + 600000 \)
\( \implies V = 628571.4285... \)
\( \implies V \approx 628571 \) Hence, the value of the machine after 6 years will be Rs. 6,28,571.
In simple words: The machine's depreciation rate is inversely proportional to \( (t+1)^2 \). We set up and solve the differential equation, using the initial value and the depreciation in the first year to find the constants 'k' and 'c'. Finally, we use the derived formula to calculate the machine's value after 6 years.

🎯 Exam Tip: Remember to include the negative sign in the differential equation for depreciation. Carefully use the given initial conditions and the value after the first year to solve for both constants of integration. Rounding should be done at the very end to maintain accuracy.