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?
Advertisements
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
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} = y\]
|
y = ax |
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
C' (x) = 2 + 0.15 x ; C(0) = 100
(1 − x2) dy + xy dx = xy2 dx
(y + xy) dx + (x − xy2) dy = 0
A population grows at the rate of 5% per year. How long does it take for the population to double?
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?
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.
Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
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).
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
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?
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).
State whether the following statement is True or False:
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
Solve the differential equation
`y (dy)/(dx) + x` = 0
