A radioactive substance disintegrates at a rate proportional to its mass. When its mass is $10$mgm, the rate of disintegration is $0.051$ mgm per day. How long will it take for the mass to be reduced from $10$mgm to $5$mgm.$[\log_{e}$$2=0.6931] 1 Answer Toolbox: • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0 • It can be written as \large\frac{g_1(y)}{g_2(y)}$$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)}$$dx+c Step 1: Let x be the mass of the substance at time t in days When t=0,x_0=10mg and \large\frac{dx}{dt}=0.051mg/day Step 2: Now \large\frac{dx}{dt}\propto x It is negative because there is decay \therefore\large\frac{dx}{dt}=$$kx$
When $t=0$
$-0.051=k(10)$
$\Rightarrow k=-0.0051$
Step 3:
$\therefore \large\frac{dx}{dt}$$=-0.0051x \large\frac{dx}{x}$$=0.0051dt$
Step 4:
Integrating on both sides we get,
$\int \large\frac{dx}{x}$$=0.0051t+\log c \log x=-0.0051t+\log c \log \large\frac{x}{c}$$=-0.0051t$
$x=ce^{\large -0.0051t}$
Step 5:
When $t=0,x_0=10=c$
$\therefore x=10e^{\large -0.0051t}$
Step 6:
When $x=5mg$,to find $t$
$5=10e^{\large 0.0051t}$
$\therefore \large\frac{1}{2}=e^{0.0051t}$