Application of differential equations
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A particular isotope of Uranium decays to half of its original amt in 25 days. Find the equation relating the amount present as a function of time and find the percentage remaining after 45 days.
Given - $fracdNdt = KN$
I used separable method to solve it to become - $ N = Ce^kt$
Let original amount be $N_0$ when $t=0$
$N_0 = C $2
Given when $N=0.5N_0$ in 25 days,
$0.5N_0 = N_0 cdot e^25k $
$k = - 0.02773$
Thus the general solution - $ N = Ce^-0.02773 t $
I need to find the particular solution using initial conditions to be able to find the amount remaining after 45 days . But I am stuck here with $C$ . Have I missed out anything ?
calculus differential-equations
asked Aug 14 at 8:34
user185692
1306
add a comment |Â
up vote
0
down vote
favorite
A particular isotope of Uranium decays to half of its original amt in 25 days. Find the equation relating the amount present as a function of time and find the percentage remaining after 45 days.
Given - $fracdNdt = KN$
I used separable method to solve it to become - $ N = Ce^kt$
Let original amount be $N_0$ when $t=0$
$N_0 = C $2
Given when $N=0.5N_0$ in 25 days,
$0.5N_0 = N_0 cdot e^25k $
$k = - 0.02773$
Thus the general solution - $ N = Ce^-0.02773 t $
I need to find the particular solution using initial conditions to be able to find the amount remaining after 45 days . But I am stuck here with $C$ . Have I missed out anything ?
calculus differential-equations
asked Aug 14 at 8:34
user185692
1306
1
You have already stated that at $t=0$ you know that $N_0 = C e^k times 0 = C$. So the full solution equals $N(t) = N_0 e^-0.02773t$, and now you can easily calculate $N(45)$
– Tim Dikland
Aug 14 at 8:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A particular isotope of Uranium decays to half of its original amt in 25 days. Find the equation relating the amount present as a function of time and find the percentage remaining after 45 days.
Given - $fracdNdt = KN$
I used separable method to solve it to become - $ N = Ce^kt$
Let original amount be $N_0$ when $t=0$
$N_0 = C $2
Given when $N=0.5N_0$ in 25 days,
$0.5N_0 = N_0 cdot e^25k $
$k = - 0.02773$
Thus the general solution - $ N = Ce^-0.02773 t $
I need to find the particular solution using initial conditions to be able to find the amount remaining after 45 days . But I am stuck here with $C$ . Have I missed out anything ?
calculus differential-equations
asked Aug 14 at 8:34
user185692
1306
A particular isotope of Uranium decays to half of its original amt in 25 days. Find the equation relating the amount present as a function of time and find the percentage remaining after 45 days.
Given - $fracdNdt = KN$
I used separable method to solve it to become - $ N = Ce^kt$
Let original amount be $N_0$ when $t=0$
$N_0 = C $2
Given when $N=0.5N_0$ in 25 days,
$0.5N_0 = N_0 cdot e^25k $
$k = - 0.02773$
Thus the general solution - $ N = Ce^-0.02773 t $
I need to find the particular solution using initial conditions to be able to find the amount remaining after 45 days . But I am stuck here with $C$ . Have I missed out anything ?
calculus differential-equations
asked Aug 14 at 8:34
user185692
1306
asked Aug 14 at 8:34
user185692
1306
asked Aug 14 at 8:34
user185692
1306
asked Aug 14 at 8:34
user185692
1306
1306
1
You have already stated that at $t=0$ you know that $N_0 = C e^k times 0 = C$. So the full solution equals $N(t) = N_0 e^-0.02773t$, and now you can easily calculate $N(45)$
– Tim Dikland
Aug 14 at 8:48
add a comment |Â
1
You have already stated that at $t=0$ you know that $N_0 = C e^k times 0 = C$. So the full solution equals $N(t) = N_0 e^-0.02773t$, and now you can easily calculate $N(45)$
– Tim Dikland
Aug 14 at 8:48
1
1
You have already stated that at $t=0$ you know that $N_0 = C e^k times 0 = C$. So the full solution equals $N(t) = N_0 e^-0.02773t$, and now you can easily calculate $N(45)$
– Tim Dikland
Aug 14 at 8:48
You have already stated that at $t=0$ you know that $N_0 = C e^k times 0 = C$. So the full solution equals $N(t) = N_0 e^-0.02773t$, and now you can easily calculate $N(45)$
– Tim Dikland
Aug 14 at 8:48
add a comment |Â
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You have already stated that at $t=0$ you know that $N_0 = C e^k times 0 = C$. So the full solution equals $N(t) = N_0 e^-0.02773t$, and now you can easily calculate $N(45)$
– Tim Dikland
Aug 14 at 8:48