Comprehensive IGNOU BMTC-132 Solved Assignment 2024 for B.Sc (G) CBCS Students

IGNOU BMTC-132 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

₹101.00

Get Instant Access

Please read the following points before ordering this IGNOU Assignment Solution.

Assignment Question Paper
Sample Solution
Frequently Asked Questions (FAQs)
Terms and Conditions

Share with your Friends

IGNOU BMTC-132 Assignment Question Paper 2024

bmtc-132-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Solved Assignment 2024

State whether the following statements are true or false. Give a short proof or a counterexample in support of your Answer:

a) A real-valued function of three variables which is continuous everywhere is differentiable.

b) The function

f (x, y) = \ln (\frac{x + y}{x})

is not homogeneous function.

c) The cylindrical coordinates of the point whose spherical coordinates is

(8, \frac{π}{6}, \frac{π}{2})

(8, \frac{π}{6}, 0)

d) The unique solution

y (x)

of an ordinary differential equation:

\frac{d y}{d x} = {\begin{cases} 0, & for x < 0 \\ 1, & for x \geq 0 \end{cases}

exists

\forall x \in R

e) The differential equation:

{[1 + {(y^{'})}^{2}]}^{\frac{5}{3}} = y^{''}

is a second order differential equation of degree 3 .

f) Differential equation:

\cos x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + x y^{2} = 0

] 0, π [

is a linear, homogeneous equation.

g) The total differential equation corresponding to the family of surfaces

x^{3} z + x^{2} y = c

, where

c

is a parameter is

3 x^{2} d z + x (y d x + x d y) = 0

h) Differential equation:

5 x^{2} y^{2} z^{2} = 2 p x^{2} y^{2} + 5 q x^{2} y^{3} + 2 p z^{2} + 9 x^{2} y^{2}

is a semi-linear partial differential equation of first order.

i) The differential equation:

v \frac{d u}{d v} = e^{2 v} + u v - u

has an integrating factor

v \exp (- v)

j) The simultaneous differential equation of simple harmonic motion of a particle in phase -plane is:

\frac{d x}{y} = \frac{d y}{- w^{2} x} = d t with y (x_{0}) = y_{0}

a) Solve the differential equation:

\frac{d y}{d x} \frac{- \tan y}{(1 + x)} = (1 + x) e^{x} \sec y

b) Show that the differential equation:

(y^{2} + y x) d x + (z^{2} + z x) d y + (y^{2} - x y) d z = 0

is integrable and find its integral.

a) Solve the differential equation:

x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} - 4 y = x^{2} + 2 \ln x

b) Using Charpit’s method, find the complete integral of the differential equation:

p (1 + q^{2}) + (b - z) q = 0,

b

being a constant.

a) Find all the first order partial derivatives of the following function:

h (x, y, t) = e^{x - t} \cos (y + t)

What is the value of

\frac{\partial h}{\partial t}

(0, \frac{π}{2}, 0)

b) Find the limit of:

f (x, y) = \frac{x^{2} - y^{2}}{x^{2} + y^{2}}

(x, y) \to (0, 0)

along:
i)

y = 3 x

ii)

y = 5 x

What can you conclude about

lim_{(x, y) \to (0, 0)} f (x, y)

? Justify your answer.

c) If

\cos α, \cos β

and

\cos γ

are the direction cosines of a line, then show that:

\sin^{2} α + \sin^{2} β + \sin^{2} γ = 2

a) Solve the differential equation:

2 x \frac{d y}{d x} + y (6 y^{2} - x - 1) = 0

b) The rate of change of the price of a commodity is proportional to the difference between the demand

D

and the supply

S

. If

D = α - b P

and

S = c \sin β t

, where

a, b, c

and

β

are constants, determine

P (t)

. It is given that at

t = 0, P = P_{0}

a) Find the differential equations of the space curve in which the two families of surfaces:

u = x^{2} + y^{2} + 3 z = c_{1}

and

v = z x + x^{2} + y^{2} = c_{2}

intersect.

b) Transform the given equation to Clairaut’s form and hence find its general solution:

x y (y - p x) = x - p y

Also find its singular solution, if it exists.

a) Find the envelope and the characteristic curves of the family of curves:

x^{2} (y - c)^{2} + z^{2} = c^{2} \cos^{2} α,

c

and

α

are constants.

b) Using Lagrange’s method, solve the differential equation:

(x^{2} - y^{2} - z^{2}) p + 2 x q = 2 x z

a) Using the method of variation of parameters, solve the differential equation:

\frac{d^{2} y}{d x^{2}} + y = \sec^{3} x

b) Solve the differential equation:

(2 x e^{y} y^{4} + 2 x y^{3} y) d x + (x^{2} y^{4} e^{y} - x^{2} y^{2} - 3 x) d y = 0

a) Show that the limit of the function $f (x, y)$ exists at the origin, where:

f (x, y) = {\begin{array}{cc} x \cos \frac{1}{y} + y \cos \frac{1}{x}, & (x, y) \neq (0, 0) \\ 0, & (x, y) = (0, 0) \end{array}

Do the repeated limits of

f (x, y)

exist? Justify your answer.

b) For the function:

f (x, y) = x^{3} + x y - 2 y^{2},

Find the polynomial given by:

f_{x x} (1, 2) (x - 2)^{2} + f_{x y} (1, 2) (x - 2) (y - 1) + f_{y y} (1, 2) (y - 1)^{2}

BMTC-132 Sample Solution 2024

bmtc-132-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Solved Assignment 2024

State whether the following statements are true or false. Give a short proof or a counterexample in support of your answer:

a) A real-valued function of three variables which is continuous everywhere is differentiable.

Answer:

The statement "A real-valued function of three variables which is continuous everywhere is differentiable" is false.

Justification

Continuity and differentiability are related but distinct properties of functions. While differentiability implies continuity, the converse is not necessarily true. A function can be continuous at a point but not differentiable at that point. This concept holds for functions of any number of variables, including functions of three variables.

Counterexample

A classic counterexample in the context of functions of a single variable is the absolute value function,

f (x) = | x |

, which is continuous everywhere but not differentiable at

x = 0

For a function of three variables, we can construct a similar example. Consider the function

f (x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}}

This function represents the distance from the origin in three-dimensional space and is continuous everywhere. However, it is not differentiable at the origin

(0, 0, 0)

To see why, consider the partial derivative with respect to

x

at the origin. The partial derivative is defined as the limit of the difference quotient:

lim_{h \to 0} \frac{f (h, 0, 0) - f (0, 0, 0)}{h}

Substituting the function, we get:

lim_{h \to 0} \frac{\sqrt{h^{2}}}{h}

This simplifies to:

lim_{h \to 0} \frac{| h |}{h}

This limit does not exist because it approaches

1

h

approaches

0

from the right and

- 1

h

approaches

0

from the left. Therefore, the function

f (x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}}

is not differentiable at the origin, even though it is continuous everywhere.

Conclusion

Thus, the statement is false. A real-valued function of three variables that is continuous everywhere is not necessarily differentiable everywhere.

b) The function

f (x, y) = \ln (\frac{x + y}{x})

is not homogeneous function.

Answer:

To determine the truth of the statement "The function

f (x, y) = \ln (\frac{x + y}{x})

is not a homogeneous function", we need to verify if the function is homogeneous of any degree. A function

f (x, y)

is homogeneous of degree

n

if it satisfies:

f (λ x, λ y) = λ^{n} f (x, y)

for all scalars

λ

. If a function satisfies this condition for some degree

n

, it is homogeneous; otherwise, it is not.

For the given function

f (x, y) = \ln (\frac{x + y}{x})

, let’s substitute

λ x

and

λ y

and simplify:

f (λ x, λ y) = \ln (\frac{λ x + λ y}{λ x})

Simplifying this expression will help us compare it with

λ^{n} f (x, y)

to determine if the function is homogeneous and, if so, of which degree. Let’s perform this calculation.

Upon simplifying the function

f (λ x, λ y) = \ln (\frac{λ x + λ y}{λ x})

, we find that it simplifies to:

f (λ x, λ y) = \ln (\frac{x + y}{x})

This is identical to the original function

f (x, y)

. Therefore, the function satisfies the condition:

f (λ x, λ y) = f (x, y)

which is the same as:

f (λ x, λ y) = λ^{0} f (x, y)

since

λ^{0} = 1

for any

λ

. This means that the function

f (x, y) = \ln (\frac{x + y}{x})

is indeed homogeneous, but of degree

n = 0

Thus, the statement "The function

f (x, y) = \ln (\frac{x + y}{x})

is not a homogeneous function" is false, as the function is homogeneous of degree 0.

c) The cylindrical coordinates of the point whose spherical coordinates is

(8, \frac{π}{6}, \frac{π}{2})

(8, \frac{π}{6}, 0)

Answer:

The statement "The cylindrical coordinates of the point whose spherical coordinates are

(8, \frac{π}{6}, \frac{π}{2})

(8, \frac{π}{6}, 0)

" is false. To demonstrate this, let’s convert the given spherical coordinates to cylindrical coordinates and compare.

Spherical Coordinates to Cylindrical Coordinates Conversion

The spherical coordinates

(r, θ, ϕ)

are related to cylindrical coordinates

(ρ, ϕ, z)

by the following relationships:

$ρ = r \sin (ϕ)$
$z = r \cos (ϕ)$
$ϕ$ (the angle in the xy-plane from the x-axis) remains the same in both coordinate systems.

Given the spherical coordinates

(8, \frac{π}{6}, \frac{π}{2})

, where

r = 8

θ = \frac{π}{6}

, and

ϕ = \frac{π}{2}

, we can convert them to cylindrical coordinates:

$ρ = 8 \sin (\frac{π}{2}) = 8$ (since $\sin (\frac{π}{2}) = 1$ )
$z = 8 \cos (\frac{π}{2}) = 0$ (since $\cos (\frac{π}{2}) = 0$ )
The angle $ϕ$ in cylindrical coordinates is the same as $θ$ in spherical coordinates, so $ϕ = \frac{π}{6}$ .

Conclusion

Therefore, the cylindrical coordinates of the point whose spherical coordinates are

(8, \frac{π}{6}, \frac{π}{2})

are

(8, \frac{π}{6}, 0)

in terms of

(ρ, ϕ, z)

, not

(8, \frac{π}{6}, 0)

in terms of

(r, θ, z)

. The statement is false because it incorrectly identifies the third component of the cylindrical coordinates as

0

instead of correctly identifying the second component as

\frac{π}{6}

and the third component as

0

d) The unique solution

y (x)

of an ordinary differential equation:

\frac{d y}{d x} = {\begin{cases} 0, & for x < 0 \\ 1, & for x \geq 0 \end{cases}

exists

\forall x \in R

Answer:

To determine whether the unique solution

y (x)

exists for the given ordinary differential equation (ODE)

\frac{d y}{d x} = {\begin{cases} 0, & for x < 0 \\ 1, & for x \geq 0 \end{cases}

for all

x \in R

, we can use the Lipschitz condition.

The Lipschitz Condition

The Lipschitz condition is a criterion used to determine the existence and uniqueness of solutions to ODEs. A function

f (x, y)

satisfies the Lipschitz condition in

y

on a domain if there exists a constant

L

such that for all

y_{1}, y_{2}

in the domain,

| f (x, y_{1}) - f (x, y_{2}) | \leq L | y_{1} - y_{2} |

Applying the Lipschitz Condition to the Given ODE

In our case, the function

f (x, y)

representing the right-hand side of the ODE is piecewise constant: it is 0 for

x < 0

and 1 for

x \geq 0

. This function does not depend on

y

, so the difference

| f (x, y_{1}) - f (x, y_{2}) |

is always 0 regardless of the values of

y_{1}

and

y_{2}

. Therefore, the function trivially satisfies the Lipschitz condition in

y

for any

x

, with any Lipschitz constant

L \geq 0

Conclusion

Since the function

f (x, y)

satisfies the Lipschitz condition, the Picard-Lindelöf theorem (or the existence and uniqueness theorem) guarantees that a unique solution

y (x)

exists for each initial value problem defined by this ODE. This holds for all

x \in R

Hence given statement is True.

e) The differential equation:

{[1 + {(y^{'})}^{2}]}^{\frac{5}{3}} = y^{''}

is a second order differential equation of degree 3 .

Answer:

The original equation is:

{[1 + {(y^{'})}^{2}]}^{\frac{5}{3}} = y^{''}

Cubing both sides gives:

{({[1 + {(y^{'})}^{2}]}^{\frac{5}{3}})}^{3} = {(y^{''})}^{3}

Simplifying the left side:

{[1 + {(y^{'})}^{2}]}^{5} = {(y^{''})}^{3}

Now, we have a second-order differential equation of degree 3 :

{(y^{''})}^{3} = {[1 + {(y^{'})}^{2}]}^{5}

Which shows that this differential equation is of second order and degree-3.

Hence, Given statement is True.

f) Differential equation:

\cos x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + x y^{2} = 0

] 0, π [

is a linear, homogeneous equation.

Answer:

The statement "The differential equation

\cos x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + x y^{2} = 0

] 0, π [

is a linear, homogeneous differential equation" is false.

Justification

To determine whether the differential equation is linear and homogeneous, let’s recall the definitions:

Linear Differential Equation: A differential equation is linear if it can be written in the form

$a_{n} (x) y^{(n)} + a_{n - 1} (x) y^{(n - 1)} + \dots + a_{1} (x) y^{'} + a_{0} (x) y = g (x)$

where $y^{(n)}$ is the $n$ th derivative of $y$ with respect to $x$ , and $a_{i} (x)$ are functions of $x$ only. The function $g (x)$ on the right-hand side is the non-homogeneous term.
Homogeneous Differential Equation: A linear differential equation is homogeneous if $g (x) = 0$ for all $x$ .

Analysis of the Given Differential Equation

The given differential equation is:

\cos x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + x y^{2} = 0

The term $\cos x \frac{d^{2} y}{d x^{2}}$ is linear because it involves the second derivative of $y$ multiplied by a function of $x$ only.
The term $\frac{d y}{d x}$ is linear because it is the first derivative of $y$ without any multiplication by $y$ or its derivatives.
However, the term $x y^{2}$ is not linear because it involves $y$ raised to the second power, which is a non-linear term.

Conclusion

Since the equation includes the non-linear term

x y^{2}

, it is not a linear differential equation. Consequently, it cannot be a homogeneous linear differential equation either. Therefore, the statement is false. The differential equation is non-linear due to the presence of the

y^{2}

term.

Get The AppClick Here

Frequently Asked Questions (FAQs)

How do I access the Complete Solution?

You can access the Complete Solution through our app, which can be downloaded using this link:

App Link

Simply click “Install” to download and install the app, and then follow the instructions to purchase the required assignment solution. Currently, the app is only available for Android devices. We are working on making the app available for iOS in the future, but it is not currently available for iOS devices.

Is this Complete Solution for IGNOU Assignments?

Yes, It is Complete Solution, a comprehensive solution to the assignments for IGNOU. Valid from January 1, 2023 to December 31, 2023.

Is the Complete Solution aligned with the IGNOU requirements?

Yes, the Complete Solution is aligned with the IGNOU requirements and has been solved accordingly.

Is the Complete Solution guaranteed error-free?

Yes, the Complete Solution is guaranteed to be error-free.The solutions are thoroughly researched and verified by subject matter experts to ensure their accuracy.

Can I access the Complete Solution anytime?

As of now, you have access to the Complete Solution for a period of 6 months after the date of purchase, which is sufficient to complete the assignment. However, we can extend the access period upon request. You can access the solution anytime through our app.

What if I need help with a specific assignment question?

The app provides complete solutions for all assignment questions. If you still need help, you can contact the support team for assistance at Whatsapp +91-9958288900

Can I access the educational materials on multiple devices?

No, access to the educational materials is limited to one device only, where you have first logged in. Logging in on multiple devices is not allowed and may result in the revocation of access to the educational materials.

How do I make a payment for accessing the solutions?

Payments can be made through various secure online payment methods available in the app.Your payment information is protected with industry-standard security measures to ensure its confidentiality and safety. You will receive a receipt for your payment through email or within the app, depending on your preference.

What are the instructions for formatting my assignments?

The instructions for formatting your assignments are detailed in the Assignment Booklet, which includes details on paper size, margins, precision, and submission requirements. It is important to strictly follow these instructions to facilitate evaluation and avoid delays.

Terms and Conditions

The educational materials provided in the app are the sole property of the app owner and are protected by copyright laws.
Reproduction, distribution, or sale of the educational materials without prior written consent from the app owner is strictly prohibited and may result in legal consequences.
Any attempt to modify, alter, or use the educational materials for commercial purposes is strictly prohibited.
The app owner reserves the right to revoke access to the educational materials at any time without notice for any violation of these terms and conditions.
The app owner is not responsible for any damages or losses resulting from the use of the educational materials.
The app owner reserves the right to modify these terms and conditions at any time without notice.
By accessing and using the app, you agree to abide by these terms and conditions.
Access to the educational materials is limited to one device only. Logging in to the app on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Our educational materials are solely available on our website and application only. Users and students can report the dealing or selling of the copied version of our educational materials by any third party at our email address (abstract4math@gmail.com) or mobile no. (+91-9958288900).

In return, such users/students can expect free our educational materials/assignments and other benefits as a bonafide gesture which will be completely dependent upon our discretion.

IGNOU BMTC-132 Solved Assignment 2024 | B.Sc (G) CBCS

IGNOU BMTC-132 Assignment Question Paper 2024

bmtc-132-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Sample Solution 2024

bmtc-132-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

Justification

Counterexample

Conclusion

Spherical Coordinates to Cylindrical Coordinates Conversion

Conclusion

The Lipschitz Condition

Applying the Lipschitz Condition to the Given ODE

Conclusion

Justification

Analysis of the Given Differential Equation

Conclusion

Frequently Asked Questions (FAQs)

Terms and Conditions

All Policies

Account

Company

Affiliate Program

Shipping Policy

Privacy Policy

Return Policy

Terms & Conditions

Cancellation Policy

Registration & Security

Refund Policy

Copyright & DMCA Notice

Privacy Policy

Content Copyright

Dispute Resolution

Terms of Services

Contact Us

IGNOU BMTC-132 Solved Assignment 2024 | B.Sc (G) CBCS

IGNOU BMTC-132 Assignment Question Paper 2024

bmtc-132-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-132 Sample Solution 2024

bmtc-132-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

Justification

Counterexample

Conclusion

Spherical Coordinates to Cylindrical Coordinates Conversion

Conclusion

The Lipschitz Condition

Applying the Lipschitz Condition to the Given ODE

Conclusion

Justification

Analysis of the Given Differential Equation

Conclusion

Frequently Asked Questions (FAQs)

Terms and Conditions

Related products

IGNOU MEG-19 Solved Assignment 2022-2023 | MEG | THE AUSTRALIAN NOVEL

IGNOU MEG-17 Solved Assignment 2022-2023 | MEG | American Drama

IGNOU MEG-05 Solved Assignment 2022-2023 | MEG | LITERARY CRITICISM & THEORY

IGNOU MST-003 Solved Assignment 2023 | PGDAST

All Policies

Account

Company

Contact Us