Advertisements
Advertisements
Question
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?
Solution
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,
\[\text{ At }t = 1990, P = 200000\text{ and at }t = 2000, P = 250000\]
\[ \therefore \log 200000 = 1990\beta + \log C . . . . . . . . \left( 2 \right) \]
\[ \log 250000 = 2000\beta + \log C . . . . . . . . . \left( 3 \right)\]
\[\text{ Subtracting }\left( 3 \right)\text{ from }\left( 2 \right), \text{ we get }\]
\[\log 200000 - \log 250000 = 10\beta\]
\[ \Rightarrow \beta = \frac{1}{10}\log\left( \frac{5}{4} \right)\]
\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right) \text{ in }\left( 2 \right),\text{ we get }\]
\[\log 200000 = 1990 \times \frac{1}{10}\log\left( \frac{5}{4} \right) + \log C\]
\[ \Rightarrow \log 200000 = 199\log\left( \frac{5}{4} \right) + \log C \]
\[ \Rightarrow \log C = \log 200000 - 199\log\left( \frac{5}{4} \right) \]
\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right), \log C = \log 200000 - 199 \log\left( \frac{5}{4} \right) \text{ and }t = 2010\text{ in }\left( 1 \right),\text{ we get }\]
\[\log\left| P \right| = \frac{1}{10} \times 2010\log \left( \frac{5}{4} \right) + \log 200000 - 199 \log\left( \frac{5}{4} \right)\]
\[ \Rightarrow \log\left| P \right| = 201 \log \left( \frac{5}{4} \right) + \log 200000 - 199\log\left( \frac{5}{4} \right)\]
\[ \Rightarrow \log\left| P \right| = \log \left( \frac{5}{4} \right)^{201} - \log \left( \frac{5}{4} \right)^{199} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^{201} \left( \frac{4}{5} \right)^{199} \right\} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^2 \right\} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left( \frac{25}{16} \times 200000 \right)\]
\[ \Rightarrow \log\left| P \right| = \log 312500\]
\[ \Rightarrow P = 312500\]
APPEARS IN
RELATED QUESTIONS
Assume that a 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.
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].
Show that y = AeBx is a solution of the differential equation
Verify that y = cx + 2c2 is a solution of the differential equation
Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[y = \left( \frac{dy}{dx} \right)^2\]
|
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
|
Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after `t` seconds.
In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
(y2 − 2xy) dx = (x2 − 2xy) dy
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?
Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\] at any point (x, y) on it.
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by
What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?
Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].
If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
The price of six different commodities for years 2009 and year 2011 are as follows:
| Commodities | A | B | C | D | E | F |
|
Price in 2009 (₹) |
35 | 80 | 25 | 30 | 80 | x |
| Price in 2011 (₹) | 50 | y | 45 | 70 | 120 | 105 |
The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x andy if the total price in 2009 is ₹ 360.
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
Solve the following differential equation.
`(x + y) dy/dx = 1`
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
`dy/dx = log x`
Choose the correct alternative:
General solution of `y - x ("d"y)/("d"x)` = 0 is
The function y = ex is solution ______ of differential equation
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0
