First order differential question

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SUMMARY

The discussion centers on solving the first-order differential equation dx/dt = 0.8x - 0.004x², which models the dissolution of KNO3 in methanol. The user initially attempts to solve the equation using separable differential equations and integration but encounters difficulties with the integration process and boundary conditions. A suggestion is made to utilize partial fractions to simplify the integration, specifically breaking down the expression 1/(0.8x - 0.004x²) into A/x + B/(0.8 - 0.004x) for easier computation. This method allows for the determination of constants A and B, facilitating the solution of the differential equation.

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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.. :frown:

so, please advise me

thanks
 
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Split into partial fractions


[tex]\frac{1}{0.8x - 0.004 x^2}=\frac{1}{x(0.8-0.004x)}=\frac{A}{x}+\frac{B}{0.8-0.004x}[/tex]


bring the right side to the same denominator, then take the numerator you get and put that equal to the numerator on the left side (1 in this case)

Now you will get an equality that is valid for all values of x, so you can pick the most convenient values of x to get A and B.

so if you had something like

A(x-1)+B(x+1)=1

to get A, you'd put x=-1 to eliminate B. To get B, you'd put x=1 and A would disappear.

Do the same with your question.
 

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