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The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is ______

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#### Solution

**e ^{–x} **

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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

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 = x_{0}

∴ log x_{0} = k × 0 + c

∴ c = `square`

∴ log x = kt + log x_{0}

∴ log x - log x_{0} = kt

∴ `log ("x"/"x"_0)`= kt ......(1)

Since the number doubles in 4 hours, i.e. when t = 4,

x = 2x_{0}

∴ `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 2^{3}

∴ `"x"/"x"_0 = 8`

∴ x = `square`

∴ number of bacteria will be 8 times the original number in 12 hours.

Bacteria increases at the rate proportional to the number of bacteria present. If the original number N doubles in 4 hours, find in how many hours the number of bacteria will be 16N.

**Solution:** Let x be the number of bacteria in the culture at time t.

Then the rate of increase of x is `("d"x)/"dt"` which is proportional to x.

∴ `("d"x)/"dt" ∝ x`

∴ `("d"x)/"dt"` = kx, where k is a constant

∴ `("d"x)/x` = kdt

On integrating, we get

`int ("d"x)/x = "k" int "dt"`

∴ log x = kt + c .....(1)

∴ x = ae^{kt} where a = e^{c }

Initially, i.e.,when t = 0, let x = N

∴ N = ae^{k(0)}

∴ a = `square`

∴ a = N, x = Ne^{kt} ......(2)

When t = 4, x = 2N

From equation (2), 2N = Ne^{4k}

∴ e^{4k }= 2

∴ e^{k }= `square`

Now we have to find out t, when x = 16N

From equation (2),

16N = Ne^{kt }

∴ 16 = e^{kt }

∴ `"t"/4 = square` hours

Hence, number of bacteria will be 16N in `square` hours

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).

The rate of decay of certain substance is directly proportional to the amount present at that instant. Initially, there are 27 gm of certain substance and 3 h later it is found that 8 gm are left, then the amount left after one more hour is ______.

The bacteria increases at the rate proportional to the number of bacteria present. If the original number 'N' doubles in 4 h, then the number of bacteria in 12 h will be ____________.

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 ______.

If r is the radius of spherical balloon at time t and the surface area of balloon changes at a constant rate K, then ______.

The population of a town increases at a rate proportional to the population at that time. If the population increases from 26,000 to 39,000 in 50 years, then the population in another 25 years will be ______ `(sqrt(3/2) = 1.225)`

The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in 7 hours, then in 35 hours its number would be ______.

The length of the perimeter of a sector of a circle is 24 cm, the maximum area of the sector is ______.

The rate of growth of bacteria is proportional to the number present. If initially, there are 1000 bacteria and the number doubles in 1 hour, the number of bacteria after `21/2` hours will be ______. `(sqrt(2) = 1.414)`

If a curve y = f(x) passes through the point (1, - 1) and satisfies the differential equation, y (1 + xy) dx = x dy, then `f(-1/2)` is equal to ______

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.

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 N be the number of bacteria present at time ‘t’.

Since the rate of increase of N is proportional to N, the differential equation can be written as –

`(dN)/dt αN`

∴ `(dN)/dt` = KN, where K is constant of proportionality

∴ `(dN)/N` = k . dt

∴ `int 1/N dN = K int 1 . dt`

∴ log N = `square` + C ...(1)

When t = 0, N = N_{0} where N_{0} is initial number of bacteria.

∴ log N_{0} = K × 0 + C

∴ C = log N_{0}

Also when t = 4, N = 2N_{0}

∴ log (2 N_{0}) = K . 4 + `square` ...[From (1)]

∴ `log((2N_0)/N_0)` = 4K,

∴ log 2 = 4K

∴ K = `square` ...(2)

Now N = ? when t = 12

From (1) and (2)

log N = `1/4 log 2 . (12) + log N_0`

log N – log N_{0} = 3 log 2

∴ `log(N_0/N_0)` = `square`

∴ N = 8 N_{0}

∴ Bacteria are increased 8 times in 12 hours.

Bacteria increase at the rate proportional to the number of bacteria present. If the original number N doubles in 3 hours, find in how many hours the number of bacteria will be 4N?

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?

Let ‘p’ be the population at time ‘t’ years.

∴ `(dp)/dt prop p`

∴ Differential equation can be written as `(dp)/dt prop kp`

where k is constant of proportionality.

∴ `(dp)/p prop k.dt`

On integrating we get

`square` = kt + c

**(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)) = (1/5log2).t`

∴ `log 4 = (1/25 log2).t`

∴ t= `square ` years