- #1
cupcake
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question :
As the salt KNO3 dissolves in methanol, the number x(t) of grams of the salt in a solution after t seconds satisfies the differential equation dx/dt = 0.8x - 0.004x^2
if x=50 when t=0, how long will it take for an additional 50g of salt to dissolve.
ok, here I'm encountering the differential part, I don't know how to solve the differential question.. so far, what I have done...
using separable differential equations..
1/(0.8x - 0.004 x^2) dx = dt
and integral both sides...
1/(0.8x - 0.004 x^2) dx = t+c
I don't know to integrate the right side, I use the integral calculator and the answer is
{ 5 ln x - 5 ln (x-200) } / 4 so, the whole equation will be
5 ln x - 5 ln (x-200) = 4 (t+c)
but, I think it doesn't work when I substitute x=50 when t=0. cause the ln (x-200) can't be minus..
so, please advise me
thanks
As the salt KNO3 dissolves in methanol, the number x(t) of grams of the salt in a solution after t seconds satisfies the differential equation dx/dt = 0.8x - 0.004x^2
if x=50 when t=0, how long will it take for an additional 50g of salt to dissolve.
ok, here I'm encountering the differential part, I don't know how to solve the differential question.. so far, what I have done...
using separable differential equations..
1/(0.8x - 0.004 x^2) dx = dt
and integral both sides...
1/(0.8x - 0.004 x^2) dx = t+c
I don't know to integrate the right side, I use the integral calculator and the answer is
{ 5 ln x - 5 ln (x-200) } / 4 so, the whole equation will be
5 ln x - 5 ln (x-200) = 4 (t+c)
but, I think it doesn't work when I substitute x=50 when t=0. cause the ln (x-200) can't be minus..
so, please advise me
thanks