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

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Replies to This Discussion

Please Discuss here about this assignment.Thanks

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

  \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

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