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प्रश्न
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is ______
उत्तर
e–x
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संबंधित प्रश्न
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.
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A body cools according to Newton’s law from 100° C to 60° C in 20 minutes. The temperature of the surrounding being 20° C. How long will it take to cool down to 30° C?
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.
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If the surrounding air is kept at 20° C and a body cools from 80° C to 70° C in 5 minutes, the temperature of the body after 15 minutes will be
Choose the correct option from the given alternatives:
If the surrounding air is kept at 20° C and a body cools from 80° C to 70° C in 5 minutes, the temperature of the body after 15 minutes will be
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 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.
A person’s assets start reducing in such a way that the rate of reduction of assets is proportional to the square root of the assets existing at that moment. If the assets at the beginning ax ‘ 10 lakhs and they dwindle down to ‘ 10,000 after 2 years, show that the person will be bankrupt in `2 2/9` years from the start.
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?
The solution of `("d"y)/("d"x) + y` = 3 is ______
The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and present population is 1 lac., when will the city have population 4,00,000?
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 = 100000
∴ log 100000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 100000
∴ log p – log 100000 = kt
∴ `log ("P"/100000)` = kt ......(i)
Since the number doubled in 25 years, i.e., when t = 25, p = 200000
∴ `log (200000/100000)` = 25k
∴ k = `square`
∴ equation (i) becomes, `log("p"/100000) = square`
When p = 400000, then find t.
∴ `log(400000/100000) = "t"/25 log 2`
∴ `log 4 = "t"/25 log 2`
∴ t = `25 (log 4)/(log 2)`
∴ t = `square` years
Find the population of city at any time t given that rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 30000 to 40000.
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" prop "p"`
∴ `"dp"/"dt"` = kp, where k is a constant.
∴ `"dp"/"p"` = k dt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e. when t = 0, let p = 30000
∴ log 30000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 30000
∴ log p - log 30000 = kt
∴ `log("p"/30000)` = kt .....(1)
when t = 40, p = 40000
∴ `log (40000/30000) = 40"k"`
∴ k = `square`
∴ equation (1) becomes, `log ("p"/30000)` = `square`
∴ `log ("p"/30000) = "t"/40 log (4/3)`
∴ p = `square`
If the population grows at the rate of 8% per year, then the time taken for the population to be doubled, is (Given log 2 = 0.6912).
If r is the radius of spherical balloon at time t and the surface area of balloon changes at a constant rate K, then ______.
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 ______
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 1,00,000, when will the city have population 4,00,000?
Let ‘p’ be the population at time ‘t’ years.
∴ `("dp")/"dt" prop "p"`
∴ Differential equation can be written as `("dp")/"dt" = "kp"`
where k is constant of proportionality.
∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
(i) Where t = 0, p = 1,00,000
∴ from (i)
log 1,00,000 = k(0) + c
∴ c = `square`
∴ log `("p"/(1,00,000)) = "kt"` ...(ii)
(ii) When t = 25, p = 2,00,000
as population doubles in 25 years
∴ from (ii) log2 = 25k
∴ k = `square`
∴ log`("p"/(1,00,000)) = (1/25log2).t`
(iii) ∴ when p = 4,00,000
`log ((4,00,000)/(1,00,000)) = (1/25log2).t`
∴ `log 4 = (1/25 log2).t`
∴ t = `square ` years
