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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 .. :)
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shams file sent on fb
Kal dun ga Solution file Jis jis ko chahye come in
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
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