मराठी

The Population of a City Increases at a Rate Proportional to the Number of Inhabitants Present at Any Time T - Mathematics

Advertisements
Advertisements

प्रश्न

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?

बेरीज

उत्तर

Let the population at any time t be P.
Given:- \[\frac{dP}{dt} \alpha P\]

\[\Rightarrow \frac{dP}{dt} = \beta P\]

\[ \Rightarrow \frac{dP}{P} = \beta dt\]

\[ \Rightarrow \log\left| P \right| = \beta t + \log C . . . . . . . . \left( 1 \right)\]

Now, 

\[\text{ At }t = 1990, P = 200000\text{ and at }t = 2000, P = 250000\]

\[ \therefore \log 200000 = 1990\beta + \log C . . . . . . . . \left( 2 \right) \]

\[ \log 250000 = 2000\beta + \log C . . . . . . . . . \left( 3 \right)\]

\[\text{ Subtracting }\left( 3 \right)\text{ from }\left( 2 \right), \text{ we get }\]

\[\log 200000 - \log 250000 = 10\beta\]

\[ \Rightarrow \beta = \frac{1}{10}\log\left( \frac{5}{4} \right)\]

\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right) \text{ in }\left( 2 \right),\text{ we get }\]

\[\log 200000 = 1990 \times \frac{1}{10}\log\left( \frac{5}{4} \right) + \log C\]

\[ \Rightarrow \log 200000 = 199\log\left( \frac{5}{4} \right) + \log C \]

\[ \Rightarrow \log C = \log 200000 - 199\log\left( \frac{5}{4} \right) \]

\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right), \log C = \log 200000 - 199 \log\left( \frac{5}{4} \right) \text{ and }t = 2010\text{ in }\left( 1 \right),\text{ we get }\]

\[\log\left| P \right| = \frac{1}{10} \times 2010\log \left( \frac{5}{4} \right) + \log 200000 - 199 \log\left( \frac{5}{4} \right)\]

\[ \Rightarrow \log\left| P \right| = 201 \log \left( \frac{5}{4} \right) + \log 200000 - 199\log\left( \frac{5}{4} \right)\]

\[ \Rightarrow \log\left| P \right| = \log \left( \frac{5}{4} \right)^{201} - \log \left( \frac{5}{4} \right)^{199} + \log 200000\]

\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^{201} \left( \frac{4}{5} \right)^{199} \right\} + \log 200000\]

\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^2 \right\} + \log 200000\]

\[ \Rightarrow \log\left| P \right| = \log\left( \frac{25}{16} \times 200000 \right)\]

\[ \Rightarrow \log\left| P \right| = \log 312500\]

\[ \Rightarrow P = 312500\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Differential Equations - Exercise 22.11 [पृष्ठ १३४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Exercise 22.11 | Q 7 | पृष्ठ १३४

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

\[\frac{d^3 x}{d t^3} + \frac{d^2 x}{d t^2} + \left( \frac{dx}{dt} \right)^2 = e^t\]

\[y\frac{d^2 x}{d y^2} = y^2 + 1\]

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]  is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]


Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex


\[\frac{dy}{dx} = \tan^{- 1} x\]


\[\cos y\frac{dy}{dx} = e^x , y\left( 0 \right) = \frac{\pi}{2}\]

Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]


\[\frac{dy}{dx} = \left( x + y \right)^2\]

\[x\frac{dy}{dx} = x + y\]

\[xy\frac{dy}{dx} = x^2 - y^2\]

\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]

 

Solve the following initial value problem:-

\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]


Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]


A population grows at the rate of 5% per year. How long does it take for the population to double?


The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.


The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by


The solution of the differential equation y1 y3 = y22 is


The differential equation satisfied by ax2 + by2 = 1 is


What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?


Which of the following differential equations has y = C1 ex + C2 ex as the general solution?


Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].


Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.


Solve the following differential equation.

y dx + (x - y2 ) dy = 0


The solution of `dy/ dx` = 1 is ______


Choose the correct alternative.

The integrating factor of `dy/dx -  y = e^x `is ex, then its solution is


A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.


`xy dy/dx  = x^2 + 2y^2`


For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0


The function y = ex is solution  ______ of differential equation


Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.


The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0


lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is


A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is


Solve the differential equation

`x + y dy/dx` = x2 + y2


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×