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Choose the correct alternative: The integrating factor of dydx+y = e–x is - Mathematics and Statistics

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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

MCQ
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उत्तर

ex 

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Application of Differential Equations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1.8: Differential Equation and Applications - Q.1

संबंधित प्रश्न

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.


The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.


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.


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


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


The normal lines to a given curve at each point (x, y) on the curve pass through (2, 0). The curve passes through (2, 3). Find the equation of the curve.


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.


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:

The solution of `dy/dx` = 1 is ______.


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


The solution of `("d"y)/("d"x) + y` = 3 is  ______


Integrating factor of `("d"y)/("d"x) + y/x` = x3 – 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


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.


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`


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 = aekt where a = e

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

∴ N = aek(0)

∴ a = `square`

∴ a = N, x = Nekt    ......(2)

When t = 4, x = 2N

From equation (2), 2N = Ne4k

∴ e4k = 2

∴  e= `square`

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

From equation (2),

16N = Nekt 

∴ 16 = ekt 

∴ `"t"/4 = square` hours

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


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 ______.


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 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)`


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.


If `(dy)/(dx)` = y + 3 > 0 and y = (0) = 2, then y (in 2) is equal to ______.


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?


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