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प्रश्न
Choose the correct alternative:
The integrating factor of `("d"y)/("d"x) + y` = e–x is
विकल्प
x
– x
ex
y = e–x
उत्तर
ex
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संबंधित प्रश्न
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after `2 1/2` hours.
[Take `sqrt2 = 1.414`]
A right circular cone has height 9 cm and radius of the base 5 cm. It is inverted and water is poured into it. If at any instant the water level rises at the rate of `(pi/"A")`cm/sec, where A is the area of the water surface A at that instant, show that the vessel will be full in 75 seconds.
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city, it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years, if the initial population is 10,000. [Take e = 2.7182]
Show that the general solution of differential equation `"dy"/"dx" + ("y"^2 + "y" + 1)/("x"^2 + "x" + 1) = 0` is given by (x + y + 1) = (1 - x - y - 2xy).
The population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousands to 60 thousands in 40 years, what will be the population in another 20 years?
(Given: `sqrt(3/2)= 1.2247)`
The rate of depreciation `(dV)/ dt` of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased ₹1,00,000 in the first year. Find its value after 6 years.
The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lac, when will the city have population 4,00,000?
If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? `("Given" sqrt(3/2) = 1.2247)`
Choose the correct alternative:
Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in
Choose the correct alternative:
The solution of `("d"y)/("d"x) + x^2/y^2` = 0 is
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is ______
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution: Let x be the number of bacteria in the culture at time t.
Then the rate of increase of x is `"dx"/"dt"` which is proportional to x.
∴ `"dx"/"dt" ∝ "x"`
∴ `"dx"/"dt"` = kx, where k is a constant
∴ `square`
On integrating, we get
`int "dx"/"x" = "k" int "dt"`
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x0
∴ log x0 = k × 0 + c
∴ c = `square`
∴ log x = kt + log x0
∴ log x - log x0 = kt
∴ `log ("x"/"x"_0)`= kt ......(1)
Since the number doubles in 4 hours, i.e. when t = 4,
x = 2x0
∴ `log ((2"x"_0)/"x"_0)` = 4k
∴ k = `square`
∴ equation (1) becomes, `log ("x"/"x"_0) = "t"/4` log 2
When t = 12, we get
`log ("x"/"x"_0) = 12/4` log 2 = 3 log 2
∴ `log ("x"/"x"_0)` = log 23
∴ `"x"/"x"_0 = 8`
∴ x = `square`
∴ number of bacteria will be 8 times the original number in 12 hours.
The population of city doubles in 80 years, in how many years will it be triple when the rate of increase is proportional to the number of inhabitants. `("Given" log3/log2 = 1.5894)`
Solution: Let p be the population at time t.
Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
∴ `"dp"/"dt" ∝ "p"`
∴ `"dp"/"dt"` = kp, where k is a constant
∴ `"dp"/"p"` = kdt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e., when t = 0, let p = N
∴ log N = k × 0 + c
∴ c = `square`
When t = 80, p = 2N
∴ log 2N = 80k + log N
∴ log 2N – log N = 80k
∴ `log ((2"N")/"N")` = 80k
∴ log (2) = 80k
∴ k = `square`
∴ p = 3N, then t = ?
∴ log p = `log2/80 "t" + log "N"`
∴ log 3N – log N = `square`
∴ t = `square` = `square` years
If the lengths of the transverse axis and the latus rectum of a hyperbola are 6 and `8/3` respectively, then the equation of the hyperbola is ______.
Let the population of rabbits surviving at a time t be governed by the differential equation `(dp(t))/dt = 1/2p(t) - 200`. If p(0) = 100, then p(t) equals ______
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, complete the following activity to find the number of times the bacteria are increased in 12 hours.
The rate of disintegration of a radioactive element at time t is proportional to its mass at that time. The original mass of 800 gm will disintegrate into its mass of 400 gm after 5 days. Find the mass remaining after 30 days.
Solution: If x is the amount of material present at time t then `dx/dt = square`, where k is constant of proportionality.
`int dx/x = square + c`
∴ logx = `square`
x = `square` = `square`.ec
∴ x = `square`.a where a = ec
At t = 0, x = 800
∴ a = `square`
At t = 5, x = 400
∴ e–5k = `square`
Now when t = 30
x = `square` × `square` = 800 × (e–5k)6 = 800 × `square` = `square`.
The mass remaining after 30 days will be `square` mg.
If `(dy)/(dx)` = y + 3 > 0 and y = (0) = 2, then y (in 2) is equal to ______.
