HSC Commerce: Marketing and Salesmanship १२ वीं कक्षा - Maharashtra State Board Important Questions

The order and degree of `((dy)/(dx))^3 - (d^3y)/(dx^3) + ye^x` = 0 are ______.

Choose the correct alternative:

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

A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution

The power of highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called ______ of the differential equation

Order and degree of differential equation`(("d"^3y)/("d"x^3))^(1/6)`= 9 is ______

State whether the following statement is True or False: 

The degree of a differential equation is the power of highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any

State whether the following statement is True or False:  

The degree of a differential equation `"e"^(-("d"y)/("d"x)) = ("d"y)/("d"x) + "c"` is not defined

State whether the following statement is True or False:

Order and degree of differential equation `x ("d"^3y)/("d"x^3) + 6(("d"^2y)/("d"x^2))^2 + y` = 0 is (2, 2)

Find the differential equation by eliminating arbitrary constants from the relation y = (c₁ + c₂x)e^x 

Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`

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`

The order and degree of the differential equation `[1 + ((dy)/(dx))^3]^(2/3) = 8((d^3y)/(dx^3))` are respectively ______.

State whether the following statement is true or false.

The integrating factor of the differential equation `(dy)/(dx) + y/x` = x³ is – x.

y² = (x + c)³ is the general solution of the differential equation ______.

Solve the following differential equation

x²y dx – (x³ + y³)dy = 0

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`.e^c

∴ x = `square`.a where a = e^c

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.

The degree of the differential equation `((d^2y)/dx^2)^2 + (dy/dx)^3` = a^x is 3.

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 = N₀ where N₀ is initial number of bacteria.

∴ log N₀ = K × 0 + C

∴ C = log N₀

Also when t = 4, N = 2N₀

∴ log (2 N₀) = 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 N₀ = 3 log 2

∴ `log(N_0/N_0)` = `square`

∴ N = 8 N₀

∴ Bacteria are increased 8 times in 12 hours.

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