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?
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,
\[\text{ At }t = 1990, P = 200000\text{ and at }t = 2000, P = 250000\]
\[ \therefore \log 200000 = 1990\beta + \log C . . . . . . . . \left( 2 \right) \]
\[ \log 250000 = 2000\beta + \log C . . . . . . . . . \left( 3 \right)\]
\[\text{ Subtracting }\left( 3 \right)\text{ from }\left( 2 \right), \text{ we get }\]
\[\log 200000 - \log 250000 = 10\beta\]
\[ \Rightarrow \beta = \frac{1}{10}\log\left( \frac{5}{4} \right)\]
\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right) \text{ in }\left( 2 \right),\text{ we get }\]
\[\log 200000 = 1990 \times \frac{1}{10}\log\left( \frac{5}{4} \right) + \log C\]
\[ \Rightarrow \log 200000 = 199\log\left( \frac{5}{4} \right) + \log C \]
\[ \Rightarrow \log C = \log 200000 - 199\log\left( \frac{5}{4} \right) \]
\[\text{ Putting }\beta = \frac{1}{10}\log \left( \frac{5}{4} \right), \log C = \log 200000 - 199 \log\left( \frac{5}{4} \right) \text{ and }t = 2010\text{ in }\left( 1 \right),\text{ we get }\]
\[\log\left| P \right| = \frac{1}{10} \times 2010\log \left( \frac{5}{4} \right) + \log 200000 - 199 \log\left( \frac{5}{4} \right)\]
\[ \Rightarrow \log\left| P \right| = 201 \log \left( \frac{5}{4} \right) + \log 200000 - 199\log\left( \frac{5}{4} \right)\]
\[ \Rightarrow \log\left| P \right| = \log \left( \frac{5}{4} \right)^{201} - \log \left( \frac{5}{4} \right)^{199} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^{201} \left( \frac{4}{5} \right)^{199} \right\} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left\{ \left( \frac{5}{4} \right)^2 \right\} + \log 200000\]
\[ \Rightarrow \log\left| P \right| = \log\left( \frac{25}{16} \times 200000 \right)\]
\[ \Rightarrow \log\left| P \right| = \log 312500\]
\[ \Rightarrow P = 312500\]
APPEARS IN
संबंधित प्रश्न
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x + y\frac{dy}{dx} = 0\]
|
\[y = \pm \sqrt{a^2 - x^2}\]
|
Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
y (1 + ex) dy = (y + 1) ex dx
\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
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.
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.
If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.
The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by
The solution of the differential equation y1 y3 = y22 is
The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
Solve the following differential equation.
x2y dx − (x3 + y3 ) dy = 0
The solution of `dy/dx + x^2/y^2 = 0` is ______
Solve the differential equation:
`e^(dy/dx) = x`
x2y dx – (x3 + y3) dy = 0
`dy/dx = log x`
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
Choose the correct alternative:
General solution of `y - x ("d"y)/("d"x)` = 0 is
An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.
Solve: ydx – xdy = x2ydx.
