# SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 chapter 1.8 - Differential Equation and Applications [Latest edition]

## Solutions for Chapter 1.8: Differential Equation and Applications

Below listed, you can find solutions for Chapter 1.8 of Maharashtra State Board SCERT Maharashtra Question Bank for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022.

Q.1Q.2Q.3Q.4Q.5Q.6
Q.1

### SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 Chapter 1.8 Differential Equation and Applications Q.1

#### MCQ [1 Mark]

Q.1 | Q 1

Choose the correct alternative:

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

• y = aex

• y = be2x

• y = be–2x

• y = eax

Q.1 | Q 2

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

• 4 hours

• 6 hours

• 8 hours

• 10 hours

Q.1 | Q 3

Choose the correct alternative:

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

• x

• – x

• ex

• y = e–x

Q.1 | Q 4

Choose the correct alternative:

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

• ye–x = x + c

• ye= x + c

• ye= 2x + c

• ye–x = 2x + c

Q.1 | Q 5

Choose the correct alternative:

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

• xy + ("d"y)/("d"x) = 0

• x ("d"y)/("d"x) + y = 0

• ("d"y)/("d"x) - 4xy =0

• x ("d"y)/("d"x) + 1 = 0

Q.1 | Q 6

Choose the correct alternative:

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

• 3log x + 7/y = c

• 2log x + 3/y = c

• log x – log y = log c

• 3log y + 2/x = c

Q.1 | Q 7

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

• 3, 1

• 1, 3

• 3, 3

• 1, 1

Q.1 | Q 8

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

• 3, 1

• 1, 3

• 3, 3

• 1, 1

Q.1 | Q 9

Choose the correct alternative:

The solution of dy/dx = 1 is ______.

• x + y = c

• xy = c

• x2 + y2 = c

• y – x = c

Q.1 | Q 10

Choose the correct alternative:

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

• x3 + y3 = 7

• x2 + y2 = c

• x3 + y3 = c

• x + y = c

Q.2

### SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 Chapter 1.8 Differential Equation and Applications Q.2

#### Fill in the following blanks [1 Mark]

Q.2 | Q 1

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

Q.2 | Q 2

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

Q.2 | Q 3

Order and degree of differential equation are always ______ integers

Q.2 | Q 4

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

Q.2 | Q 5

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

Q.2 | Q 6

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

Q.2 | Q 7

Integrating factor of ("d"y)/("d"x) + y/x = x3 – 3 is ______

Q.2 | Q 8

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

Q.2 | Q 9

The function y = ex is solution  ______ of differential equation

Q.2 | Q 10

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

Q.3

### SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 Chapter 1.8 Differential Equation and Applications Q.3

#### [1 Mark]

Q.3 | Q 1

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

• True

• False

Q.3 | Q 2

State whether the following statement is True or False:

Order and degree of differential equation are always positive integers.

• True

• False

Q.3 | Q 3

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

• True

• False

Q.3 | Q 4

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

• True

• False

Q.3 | Q 5

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

• True

• False

Q.3 | Q 6

State whether the following statement is True or False:

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

• True

• False

Q.3 | Q 7

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)

• True

• False

Q.3 | Q 8

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.

• True

• False

Q.3 | Q 9

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

• True

• False

Q.3 | Q 10

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

• True

• False

Q.4

### SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 Chapter 1.8 Differential Equation and Applications Q.4

#### Attempt the following questions [3 Marks]

Q.4 | Q 1

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

Q.4 | Q 2

Solve the following differential equation:

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

Q.4 | Q 3

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

Q.4 | Q 4

Solve the differential equation xdx + 2ydy = 0

Q.4 | Q 5

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

Q.4 | Q 6

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

Q.4 | Q 7

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

Q.4 | Q 8

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

Q.4 | Q 9

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

Q.4 | Q 10

Solve: ("d"y)/("d"x) + 2/xy = x2

Q.5

### SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 Chapter 1.8 Differential Equation and Applications Q.5

#### Attempt the following questions [4 Marks]

Q.5 | Q 1

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

Q.5 | Q 2

Solve the following differential equation:

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

Q.5 | Q 3

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

Q.5 | Q 4

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)

Q.5 | Q 5

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)

Q.5 | Q 6

Solve the following differential equation

yx ("d"y)/("d"x) = x2 + 2y2

Q.5 | Q 7

Solve the following differential equation

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

Q.5 | Q 8

For the differential equation, find the particular solution

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

Q.5 | Q 9

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

Q.5 | Q 10

Solve the following differential equation

x^2  ("d"y)/("d"x) = x2 + xy − y2

Q.6

### SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 Chapter 1.8 Differential Equation and Applications Q.6

#### Attempt the following questions (Activity) [4 Marks]

Q.6 | Q 1

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 = cex is the required general solution of the given differential equation

Q.6 | Q 2

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

Q.6 | Q 3

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

Q.6 | Q 4

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.

Q.6 | Q 5

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

Q.6 | Q 6

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.

Q.6 | Q 7

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

Q.6 | Q 8

Find the particular solution of the following differential equation

("d"y)/("d"x) = e2y cos x, when x = pi/6, y = 0.

Solution: The given D.E. is ("d"y)/("d"x) = e2y 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 = – 2c

This is general solution.

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

"e"^0 + 2sin  pi/6 = c

∴ c = square

∴ particular solution is square

Q.6 | Q 9

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

Q.6 | Q 10

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

## Solutions for Chapter 1.8: Differential Equation and Applications

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## SCERT Maharashtra Question Bank solutions for 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 chapter 1.8 - Differential Equation and Applications

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Concepts covered in 12th Standard HSC Mathematics and Statistics (Commerce) Maharashtra State Board 2022 chapter 1.8 Differential Equation and Applications are Differential Equations, Order and Degree of a Differential Equation, Formation of Differential Equation by Eliminating Arbitary Constant, Differential Equations with Variables Separable Method, Homogeneous Differential Equations, Linear Differential Equations, Application of Differential Equations.

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