# Radium disappears at a rate propotional to the amount present .If $5\%$ of the original amount disappears in $50$ years , how much will remain at the end of $100$ years [Take $A_{0}$ as a intial amount.]

Toolbox:
• First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
• It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
• The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
Let $x$ be the quantity of radium at time $t$ (in years)
At $t=0$,the quantity was $A_0$
Step 2:
The rate of change of $x,\large\frac{dx}{dt}\propto x$
$\therefore \large\frac{dx}{dt}$$=kx Where k is the constant of proportionality . \large\frac{dx}{x}$$=kdt$
Step 3:
Integrating on both sides we get,
$\int\large\frac{dx}{x}$$=\int kdt+\log c \log x=kt+\log c \log \large\frac{x}{c}$$=kt$
$\Rightarrow x=ce^{kt}$
Step 4:
When $t=0,x=A_0$
$\therefore A_0=c$
When $t=50,x=0.95A_0$
$\therefore 0.95A_0=A_0e^{50k}$
$\Rightarrow e^{50k}=0.95$
Step 5:
When $t=100$
$x=A_0e^{100k}$
$\;\;\;=A_0e^{50k}.e^{50k}$
$\;\;\;=A_0(0.95)(0.95)$
$\;\;\;=0.9025A_0$
$\therefore$ the amount remaining after 100 years =$0.9025A_0$