SCERT Maharashtra solutions for Mathematics and Statistics (Commerce) [English] 12 Standard HSC chapter 1.8 - Differential Equation and Applications [Latest edition]

Choose the correct alternative:

Solution of the equation `x("d"y)/("d"x)` = y log y is

Choose the correct alternative:

Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in

Choose the correct alternative:

The integrating factor of `("d"y)/("d"x) + y` = e^–x is

Choose the correct alternative:

The integrating factor of `("d"^2y)/("d"x^2) - y` = e^x, is e^–x, then its solution is

Choose the correct alternative:

Differential equation of the function c + 4yx = 0 is

Choose the correct alternative:

General solution of `y - x ("d"y)/("d"x)` = 0 is

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

Choose the correct alternative:

The order and degree of `(1 + (("d"y)/("d"x))^3)^(2/3) = 8 ("d"^3y)/("d"x^3)` are respectively

Choose the correct alternative:

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

Choose the correct alternative:

The solution of `("d"y)/("d"x) + x^2/y^2` = 0 is

Order of highest derivative occurring in the differential equation is called the ______ of the differential equation

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

Order and degree of differential equation are always ______ integers

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

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 ______

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

The function y = e^x is solution  ______ of differential equation

The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 is ______

State whether the following statement is True or False:

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

State whether the following statement is True or False: 

Order and degree of differential equation are always positive integers.

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

 Order of highest derivative occurring in the differential equation is called the degree of the differential equation

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:   

A homogeneous differential equation is solved by substituting y = vx and integrating it

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)

State whether the following statement is True or False:

Number of arbitrary constant in the general solution of a differential equation is equal to order of D.E.

State whether the following statement is True or False:

A differential equation in which the dependent variable, say y, depends only on one independent variable, say x, is called as ordinary differential equation

The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`

Solve the differential equation `("d"y)/("d"x) + y` = e^−x 

Solve the following differential equation:

`"x" "dy"/"dx" + "2y" = "x"^2 * log "x"`

Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`

Solve the differential equation xdx + 2ydy = 0

Solve the differential equation (x² – yx²)dy + (y² + xy²)dx = 0

Solve the following differential equation `("d"y)/("d"x)` = x²y + y

Find the differential equation by eliminating arbitrary constants from the relation x² + y² = 2ax

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

Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0

Solve: `("d"y)/("d"x) + 2/xy` = x² 

For the differential equation, find the particular solution (x – y²x) dx – (y + x²y) dy = 0 when x = 2, y = 0

Solve the following differential equation:

`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`

Form the differential equation from the relation x² + 4y² = 4b²

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)`

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 `5/2` hours  `("Given"  sqrt(2) = 1.414)`

Solve the following differential equation

`yx ("d"y)/("d"x)` = x² + 2y² 

Solve the following differential equation

`y log y ("d"x)/("d"y) + x` = log y

For the differential equation, find the particular solution

`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0

Solve the following differential equation y²dx + (xy + x²) dy = 0

Solve the following differential equation

`x^2  ("d"y)/("d"x)` = x² + xy − y² 

Find the general solution of the equation `("d"y)/("d"x) - y` = 2x.

Solution: The equation `("d"y)/("d"x) - y` = 2x

is of the form `("d"y)/("d"x) + "P"y` = Q

where P = `square` and Q = `square`

∴ I.F. = `"e"^(int-"d"x)` = e^–x

∴ the solution of the linear differential equation is

ye^–x = `int 2x*"e"^-x  "d"x + "c"`

∴ ye^–x  = `2int x*"e"^-x  "d"x + "c"`

= `2{x int"e"^-x "d"x - int square  "d"x* "d"/("d"x) square"d"x} + "c"`

= `2{x ("e"^-x)/(-1) - int ("e"^-x)/(-1)*1"d"x} + "c"`

∴ ye^–x = `-2x*"e"^-x + 2int"e"^-x "d"x + "c"`

∴ e^–xy = `-2x*"e"^-x+ 2 square + "c"`

∴ `y + square + square` = ce^x is the required general solution of the given differential equation

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`

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`

Solve the following differential equation 

sec² x tan y dx + sec² y tan x dy = 0

Solution: sec² x tan y dx + sec² y tan x dy = 0

∴ `(sec^2x)/tanx  "d"x + square` = 0

Integrating, we get

`square + int (sec^2y)/tany  "d"y` = log c

Each of these integral is of the type

`int ("f'"(x))/("f"(x))  "d"x` = log |f(x)| + log c

∴ the general solution is

`square + log |tan y|` = log c

∴ log |tan x . tan y| = log c

`square`

This is the general solution.

Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c

Find the particular solution of the following differential equation

`("d"y)/("d"x)` = e^2y cos x, when x = `pi/6`, y = 0.

Solution: The given D.E. is `("d"y)/("d"x)` = e^2y cos x

∴ `1/"e"^(2y)  "d"y` = cos x dx

Integrating, we get

`int square  "d"y` = cos x dx

∴ `("e"^(-2y))/(-2)` = sin x + c₁

∴ e^–2y = – 2sin x – 2c₁

∴ `square` = c, where c = – 2c₁ 

This is general solution.

When x = `pi/6`, y = 0, we have

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `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 = 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

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