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?
Advertisements
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
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Show that y = AeBx is a solution of the differential equation
Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
(1 + x2) dy = xy dx
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
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.
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.
A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.
Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\] = x (x + 1) and passing through (1, 0).
Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.
Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
`y=sqrt(a^2-x^2)` `x+y(dy/dx)=0`
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
For each of the following differential equations find the particular solution.
`y (1 + logx)dx/dy - x log x = 0`,
when x=e, y = e2.
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
Choose the correct alternative.
The differential equation of y = `k_1 + k_2/x` is
Choose the correct alternative.
The solution of `x dy/dx = y` log y is
Solve the differential equation:
dr = a r dθ − θ dr
`dy/dx = log x`
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve the differential equation xdx + 2ydy = 0
Solve: `("d"y)/("d"x) + 2/xy` = x2
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is ______.
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
