IGNOU MMT-007 Solved Assignment 2024 for M.Sc. MACS

IGNOU MMT-007 Solved Assignment 2024 | M.Sc. MACS

Q: 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.

Q: 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.

Q: Is the Complete Solution aligned with the IGNOU requirements?

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

Q: 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.

Q: 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.

Q: 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

Q: 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.

Q: 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.

Q: 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.

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

₹365.00

Get Instant Access

Access via our Android App Only

Please read the following points before ordering :

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

Share with your Friends

IGNOU MMT-007 Assignment Question Paper 2024

mmt-007-solved-assignment-2024-qp-23dc1574-c435-4730-a3b0-c1df719cc7c8

a) Show that $f (x, y) = x y$
i) satisfies a Lipschitz condition on any rectangle $a \leq x \leq b$ and $c \leq y \leq d$ ;
ii) satisfies a Lipschitz condition on any strip $a \leq x \leq b$ and $- \infty < y < \infty$ ;
iii) does not satisfy a Lipschitz condition on the entire plane.
b) Use Frobenious method to find the series solution about $x = 0$ of the equation

x (1 - x) \frac{d^{2} y}{d x^{2}} - (1 + 3 x) \frac{d y}{d x} - y = 0 .

a) For the following differential equation locate and classify its singular points on the $x$ -axis
i) $x^{3} (x - 1) y^{''} - 2 (x - 1) y^{'} + 3 x y = 0$
ii) $(3 x + 1) x y^{''} - (x + 1) y^{'} + 2 y = 0$
b) Show that $L_{n + 1} (x) = (2 n + 1 - x) L_{n} (x) - n^{2} L_{n - 1} (x)$ .
c) Show that $\int_{- 1}^{1} x^{2} P_{n - 1} (x) P_{n + 1} (x) d x = \frac{2 n (n + 1)}{(2 n - 1) (2 n + 1) (2 n + 3)}$ .
d) Construct Green’s function for the differential equation

x y^{''} + y^{'} = 0, 0 < x < ℓ

under the conditions that

y (0)

is bounded and

y (ℓ) = 0

.
3. a) Show that between every successive pair of zeros of

J_{0} (x)

there exists a zero of

J_{1} (x)

.
b) Using the transformation

y = x^{1 / 2} u, 2 x^{3 / 2} = 3 z

find the solution of the equation

y^{''} + x y = 0

in terms of Bessel’s functions.
c) Show that

\int_{0}^{\infty} e^{- a x} J_{0} (b x) d x = \frac{1}{\sqrt{a^{2} + b^{2}}}, a > 0 b > 0

.
4. a) Find the Laplace transform of

\frac{\cos \sqrt{t}}{\sqrt{t}}

.
b) If

k_{m}

and

k_{n}

are distinct roots of Bessel function

J_{p} (k b) = 0

with

p \geq 0, b > 0

then show that

\int_{0}^{b} x J_{p} (k_{m} x) J_{p} (k_{n} x) d x = {\begin{cases} 0 & if & m \neq n \\ \frac{b^{2}}{2} [J_{p + 1} (k_{n} b)] & if & m = n \end{cases} .

a) Solve the following IBVP using the Laplace transform technique:

\begin{aligned} u_{t} = u_{x x}, 0 < x < 1, t > 0 . \\ u (0, t) = 1, u (1, t) = 1, t > 0 \\ u (x, 0) = 1 + \sin π x, 0 < x < 1 . \end{aligned}

b) If the Fourier cosine transform of

f (x)

α^{n} e^{- a α}

, then show that

f (x) = \frac{2}{π} \frac{n! \cos (n + 1) θ}{{(a^{2} + x^{2})}^{\frac{n + 1}{2}}} .

a) Find the displacement $u (x, t)$ of an infinite string using the method of Fourier transform given that the string is initially at rest and that the initial displacement is $f (x), - \infty < x < \infty$ .
b) Using Fourier integral representation show that

\int_{0}^{\infty} \frac{\cos (α x) + α \sin (α x)}{1 + α^{2}} d α = {\begin{array}{ccc} 0 & if & x < 0 \\ π / 2 & if & x = 0 . \\ π e^{- x} & if & x > 0 \end{array}

a) Using Runge-Kutta second order method with
(i) $h = 0.1$ , (ii) $h = 0.2$ , solve the initial value problem

y^{'} = y^{2} \sin x, y (0) = 1 .

Upto

x = 0.4

. If the exact solution is

y = \sec x

, obtain the error.
b) Solve the heat conduction equation

u_{t} = u_{x x}

in the region

R (0 \leq x \leq 1, t > 0)

with the initial and boundary conditions

u (x, 0) = 0, u (0, t) = 0, u (1, t) = t

using Crank-Nicolson method with

h = 0.25

and

λ = 1

upto two time steps.
8. a) Using second order finite Difference method, solve the boundary value problem

y^{''} + 5 y^{'} + 4 y = 1, y (0) = 0, y (1) = 0, h = 1 / 4 .

b) Solve the wave equation

u_{t t} = u_{x x}

with the initial and boundary conditions

u (x, 0) = 0, u_{t} (x, 0) = 0, u (0, t) = 0, u (1, t) = 100 \sin π t .

with

h = k = 0.25

, using the explicit method upto four time levels.
9. a) Find an approximate value of

y (1.0)

for the initial value problem

y^{'} = x^{3} - y^{3}, y (0) = 1

using the multiple method

y_{n + 1} = y_{n} + \frac{h}{3} [7 f_{n} - 2 f_{n - 1} + f_{n - 2}]

with step length

h = 0.2

. Calculate the starting values using Runge-Kutta second order method with the same

h

.
b) Using standard five point formula, solve the Laplace equation

\nabla^{2} u = 0

R

where

R

is the square

0 \leq x \leq 1, 0 \leq y \leq 1

subject to the boundary conditions

u (x, y) = x^{2} - y^{2}

x = 0, y = 0, y = 1

and

3 u + 2 \frac{\partial u}{\partial x} = x^{2} + y^{2}

x = 1

. Assume

h = k = 1 / 2

.
10. a) Find an approximate value of

y (1.0)

for the initial value problem

y^{'} = x - 2 y, y (0) = 1

using Milne-Simpson’s method

y_{n + 1} = y_{n - 1} + \frac{h}{3} [f_{n + 1} + 4 f_{n} + f_{n - 1}]

with the step length

h = 0.2

. Calculate the starting value using Runge-Kutta fourth order method with the same

h

.
b) Using fourth order Taylor series method with

h = 0.2

, solve the initial value problem

y^{'} = x + \cos y, y (0) = 0

upto

x = 1

MMT-007 Sample Solution 2024

mmt-007-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

a) Show that $f (x, y) = x y$
i) satisfies a Lipschitz condition on any rectangle $a \leq x \leq b$ and $c \leq y \leq d$ ;

ii) satisfies a Lipschitz condition on any strip

a \leq x \leq b

and

- \infty < y < \infty

;

iii) does not satisfy a Lipschitz condition on the entire plane.

Answer:

To show that the function

f (x, y) = x y

satisfies a Lipschitz condition on certain domains, we need to find a constant

L

such that for all points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

in the domain,

| f (x_{1}, y_{1}) - f (x_{2}, y_{2}) | \leq L \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}

i) On a rectangle $a \leq x \leq b$ and $c \leq y \leq d$ :

Consider any two points

(x_{1}, y_{1})

and

(x_{2}, y_{2})

in the rectangle. Then,

\begin{aligned} | f (x_{1}, y_{1}) - f (x_{2}, y_{2}) | & = | x_{1} y_{1} - x_{2} y_{2} | \\ = | x_{1} y_{1} - x_{1} y_{2} + x_{1} y_{2} - x_{2} y_{2} | \\ \leq | x_{1} | | y_{1} - y_{2} | + | y_{2} | | x_{1} - x_{2} | \\ \leq max {| x_{1} |, | y_{2} |} (| x_{1} - x_{2} | + | y_{1} - y_{2} |) \end{aligned}

Since

a \leq x \leq b

and

c \leq y \leq d

, we can take

L = max {| a |, | b |, | c |, | d |}

. Then,

| f (x_{1}, y_{1}) - f (x_{2}, y_{2}) | \leq L (| x_{1} - x_{2} | + | y_{1} - y_{2} |) \leq L \sqrt{2} \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}

So,

f (x, y) = x y

satisfies a Lipschitz condition on any rectangle

a \leq x \leq b

and

c \leq y \leq d

with Lipschitz constant

L \sqrt{2}

ii) On a strip $a \leq x \leq b$ and $- \infty < y < \infty$ :

Using a similar argument as in part (i), we can show that

f (x, y) = x y

satisfies a Lipschitz condition on any strip

a \leq x \leq b

and

- \infty < y < \infty

with a Lipschitz constant

L = max {| a |, | b |}

iii) On the entire plane:

To show that

f (x, y) = x y

does not satisfy a Lipschitz condition on the entire plane, consider the points

(x_{1}, y_{1}) = (1, n)

and

(x_{2}, y_{2}) = (0, 0)

for some large

n \in N

. Then,

| f (x_{1}, y_{1}) - f (x_{2}, y_{2}) | = | 1 \cdot n - 0 \cdot 0 | = n

However,

\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} = \sqrt{1^{2} + n^{2}} \approx n

for large

n

. If

f

were Lipschitz on the entire plane, there would exist a constant

L

such that

n \leq L \cdot n

for all

n

, which is impossible. Therefore,

f (x, y) = x y

does not satisfy a Lipschitz condition on the entire plane.

b) Use Frobenious method to find the series solution about

x = 0

of the equation

x (1 - x) \frac{d^{2} y}{d x^{2}} - (1 + 3 x) \frac{d y}{d x} - y = 0 .

Answer:

To solve the differential equation using the Frobenius method, we assume a solution of the form:

y = \sum_{n = 0}^{\infty} a_{n} x^{n + r}

where

r

is a constant to be determined. We then substitute this series into the differential equation and solve for the coefficients

a_{n}

First, we calculate the derivatives:

\frac{d y}{d x} = \sum_{n = 0}^{\infty} (n + r) a_{n} x^{n + r - 1}

\frac{d^{2} y}{d x^{2}} = \sum_{n = 0}^{\infty} (n + r) (n + r - 1) a_{n} x^{n + r - 2}

Substituting these into the differential equation, we get:

x (1 - x) \sum_{n = 0}^{\infty} (n + r) (n + r - 1) a_{n} x^{n + r - 2} - (1 + 3 x) \sum_{n = 0}^{\infty} (n + r) a_{n} x^{n + r - 1} - \sum_{n = 0}^{\infty} a_{n} x^{n + r} = 0

Expanding and combining like terms, we have:

\sum_{n = 0}^{\infty} [(n + r) (n + r - 1) - (n + r) + 1] a_{n} x^{n + r} - \sum_{n = 0}^{\infty} [3 (n + r) + (n + r) (n + r - 1)] a_{n} x^{n + r + 1} = 0

Equating coefficients of

x^{n + r}

and

x^{n + r + 1}

to zero, we get the recurrence relations:

(n + r) (n + r - 2) a_{n} - (3 (n + r) + (n + r) (n + r - 1)) a_{n + 1} = 0

Simplifying, we find:

(n + r) (n + r - 2) a_{n} - (n + r + 1) (n + r + 3) a_{n + 1} = 0

For

n = 0

, we have the indicial equation:

r (r - 2) = 0

which gives

r = 0

r = 2

For

r = 0

, the recurrence relation becomes:

n (n - 2) a_{n} - (n + 1) (n + 3) a_{n + 1} = 0

a_{n + 1} = \frac{n (n - 2)}{(n + 1) (n + 3)} a_{n}

Since

a_{0}

is arbitrary, let’s choose

a_{0} = 1

for simplicity. Then, we can find the next few coefficients:

For

n = 0

a_{1} = \frac{0 (0 - 2)}{(0 + 1) (0 + 3)} a_{0} = 0

For

n = 1

a_{2} = \frac{1 (1 - 2)}{(1 + 1) (1 + 3)} a_{1} = 0

For

n = 2

a_{3} = \frac{2 (2 - 2)}{(2 + 1) (2 + 3)} a_{2} = 0

For

n = 3

a_{4} = \frac{3 (3 - 2)}{(3 + 1) (3 + 3)} a_{3} = 0

And so on. It appears that all coefficients after

a_{0}

are zero. Therefore, the solution for

r = 0

is:

y = a_{0} = 1

Now, let’s consider the case

r = 2

For

r = 2

, the recurrence relation becomes:

(n + 2) (n) a_{n} - (n + 3) (n + 5) a_{n + 1} = 0

a_{n + 1} = \frac{(n + 2) (n)}{(n + 3) (n + 5)} a_{n}

Again, let’s choose

a_{0} = 1

. Then, we can find the next few coefficients:

For

n = 0

a_{1} = \frac{2 (0)}{(0 + 3) (0 + 5)} a_{0} = 0

For

n = 1

a_{2} = \frac{3 (1)}{(1 + 3) (1 + 5)} a_{1} = 0

And so on. It appears that all coefficients after

a_{0}

are also zero in this case. Therefore, the solution for

r = 2

is:

y = a_{0} x^{2} = x^{2}

In conclusion, the series solutions about

x = 0

of the equation

x (1 - x) \frac{d^{2} y}{d x^{2}} - (1 + 3 x) \frac{d y}{d x} - y = 0

are

y = 1

and

y = x^{2}

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.

Categories IGNOU Assignment Solution, M.Sc. MACS, M.Sc. MACS 2024 Tags "IGNOU, 2024, Academic Success, Assignment Solutions, IGNOU Academic Guides 2024, IGNOU Mathematics Assignments, IGNOU MMT-007 Solved Assignment 2024, M.Sc. MACS, M.Sc. MACS Assignment Solutions, Mathematics, MMT-007, MMT-007 Assignment Answers, Solved Assignment, study materials

IGNOU MMT-007 Solved Assignment 2024 | M.Sc. MACS

IGNOU MMT-007 Assignment Question Paper 2024

mmt-007-solved-assignment-2024-qp-23dc1574-c435-4730-a3b0-c1df719cc7c8

MMT-007 Sample Solution 2024

mmt-007-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

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 MMT-007 Solved Assignment 2024 | M.Sc. MACS

IGNOU MMT-007 Assignment Question Paper 2024

mmt-007-solved-assignment-2024-qp-23dc1574-c435-4730-a3b0-c1df719cc7c8

MMT-007 Sample Solution 2024

mmt-007-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

Frequently Asked Questions (FAQs)

Terms and Conditions

Related products

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

IGNOU MEG-16 Solved Assignment 2022-2023 | MEG | Indian Folk Literature

IGNOU MEG-06 Solved Assignment 2022-2023 | MEG | AMERICAN LITERATURE

IGNOU MEG-03 Solved Assignment 2022-2023 | MEG | British Novel

All Policies

Account

Company

Contact Us