We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>

www.vustudents.ning.com

 www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More

Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion

MTH401 Differential Equations 3rd assignment from 09-12 lectures have been uploaded on VULMS. Its due date is Tuesday, May 24, 2016.

+ How to Join Subject Study Groups & Get Helping Material?

+ How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators?

+ VU Students Reserves The Right to Delete Your Profile, If?

Views: 1633

.

+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)

+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)

Attachments:

Replies to This Discussion

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Have Any Body got an Idea how to solve it..??

you give me sight of this problem statement i will get you through it thankyou !!

i got the  sol anyone please check if thats right or wrong !! plzz ridllet3@gmail.com mine id text me with mth401

ok show me your solution text me .. :)

shams file sent on fb

Kal dun ga Solution file Jis jis ko chahye come in

ider na pic upload ho rhe na file warna me ider he de deta agr aap k pass sol he to plzz de dean

sol mth401 verify it please and then comment here if there is any fault

sol iss ko mathype pe repaste kegye ga

mill jae ga sol

$\begin{gathered} \frac{{dA(t)}}{{dt}} = {k_1}[M - A(t)] - {k_2}A(t) \hfill \\ = {k_1}M - {k_1}A(t) - {k_2}A(t) \hfill \\ = {k_1}M - A(t)[{k_1} + {k_2}] \hfill \\ dt = \frac{{dA(t)}}{{{k_1}M - A(t)[{k_1} + {k_2}]}} \hfill \\ \int {dt = \int {\frac{{dA(t)}}{{{k_1}M - A(t)[{k_1} + {k_2}]}}} } \hfill \\ t + C = \frac{{ - \ln [M{k_1} - At({k_1} - {k_2})]}}{{{k_1} - {k_2}}} + C \hfill \\ t = \frac{{ - \ln [M{k_1} - At({k_1} - {k_2})]}}{{{k_1} - {k_2}}} \hfill \\ \ln [M{k_1} - At({k_1} - {k_2})] = - ({k_1} + {k_2})t \hfill \\ [M{k_1} - At({k_1} - {k_2})] = {e^{ - ({k_1} + {k_2})t}} \hfill \\ - At({k_1} - {k_2}) = {e^{ - ({k_1} + {k_2})t}} - M{k_1} \hfill \\ A = \frac{{{e^{ - ({k_1} + {k_2})t}} - M{k_1}}}{{t({k_1} - {k_2})}} \hfill \\ A = 0 \hfill \\ putting value in equation we have 0=0 Thus proved A(0)=0 Now as time goes on \frac{{{e^{ - ({k_1} + {k_2})t}} - M{k_1}}}{{t({k_1} - {k_2})}} = 0 \hfill \\ M{k_1} = {e^{ - ({k_1} + {k_2})t}} \hfill \\ {\lim _{t - > \infty }}\frac{{{e^{ - ({k_1} + {k_2})t}}}}{{t({k_1} - {k_2})}} = \frac{{{e^\infty }}}{\infty } = \frac{1}{\infty } = 0 \hfill \\ \end{gathered}$

This means that relatively quickly, the amount memorized becomes constant. That constant is always less than the total amount to be memorized

not responding Hamza bro

Latest Activity

Marium Batool updated their profile
7 minutes ago
Hafiza Iqra Khan liked ٢١ دن's discussion گناہ
45 minutes ago
1 hour ago
Muhammad Irfan joined + M.Tariq Malik's group

CS506 Web Design and Development

1 hour ago
Muhammad Irfan joined + M.Tariq Malik's group

CS510 Software Requirements and Specifications

1 hour ago
1 hour ago
Muhammad Irfan and Areeha Chuhdary joined + M.Tariq Malik's group

STA301 Statistics and Probability

1 hour ago
Muhammad Irfan joined + M.Tariq Malik's group

MGT502 Organizational Behavior

1 hour ago
1 hour ago
1 hour ago
Fari Malik liked Zain Kazimi's discussion Self Poetry Dedicated to Nasir Kazimi
2 hours ago

1

2