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प्रश्न
Choose the correct alternative:
The solution of `("d"y)/("d"x) + x^2/y^2` = 0 is
विकल्प
x3 + y3 = 7
x2 + y2 = c
x3 + y3 = c
x + y = c
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उत्तर
x3 + y3 = c
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संबंधित प्रश्न
If the population of a country doubles in 60 years, in how many years will it be triple (treble) under the assumption that the rate of increase is proportional to the number of inhabitants?
(Given log 2 = 0.6912, log 3 = 1.0986)
The rate of disintegration of a radioactive element at any time t is proportional to its mass at that time. Find the time during which the original mass of 1.5 gm will disintegrate into its mass of 0.5 gm.
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.
Radium decomposes at the rate proportional to the amount present at any time. If p percent of the amount disappears in one year, what percent of the amount of radium will be left after 2 years?
Choose the correct option from the given alternatives:
The decay rate of certain substances is directly proportional to the amount present at that instant. Initially there are 27 grams of substance and 3 hours later it is found that 8 grams left. The amount left after one more hour is
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).
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 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.
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 integrating factor of `("d"^2y)/("d"x^2) - y` = ex, is e–x, then its solution is
Choose the correct alternative:
The solution of `dy/dx` = 1 is ______.
Integrating factor of `("d"y)/("d"x) + y/x` = x3 – 3 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.
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 = ec
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
∴ ek = `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 equation of tangent at P(- 4, - 4) on the curve x2 = - 4y 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 length of the perimeter of a sector of a circle is 24 cm, the maximum area of the sector is ______.
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 ______.
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 1,00,000, 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
