Differential equations HWK?

Question

Please could anyone help me with this HWK??

At time t minutes, a solution contains x grams of a substance S. Because of a chemical reaction, x increases at a rate per minute given by

dx/dt= (6-x)(3-x)

At one moment, there are 0.5g of S in the solution. Calculate the time, innseconds, that must elapse before there will be 2.5 grams of S in the solution.

Any help would be apppreciated with an explanation if poss! P.s This may have something to do with integration by partial fractions, as i managed to do th epart before it using this method.

Answer 1

You must integrate dx/dt obtaining F(x).
Your answer is F(2.5)-F(0.5)

PS: s_h_mc bellow is right

Answer 2

The above answerers have at a loss for words the particular answer with the particular vital, those words have diverse meanings and are by making use of no skill interchangeable. detect the complementary function by skill of changing the auxiliary equation: 2(d²y / dx²) + dy / dx - y = 0 2m² + m - a million = 0 (2m - a million)(m + a million) = 0 2m - a million = 0 OR m + a million = 0 2m = a million OR m = -a million m = ½ OR m = -a million y? = C??^(x / 2) + C???? y? = C??^(x / 2) + C? / ?? detect the particular vital by skill of comparing coefficients: y? = Ax + B dy? / dx = A d²y? / dx² = 0 2(d²y? / dx²) + dy? / dx - y? = x A - (Ax + B) = x A - Ax - B = x (A - B) - Ax = x A = -a million A - B = 0 B = A B = -a million y? = -x - a million detect the common answer by skill of mixing those 2 factors: y = y? + y? y = C??^(x / 2) + C? / ?? - x - a million

Answer 3

Uh, yeah...

Yes, you need to integrate by partial fractions.
I came up with (-1/3)/(6-x) + (1/3)/(3-x) dx = dt
Integrate both sides:
(1/3)ln(6-x) - (1/3)ln(3-x) = t + C.

So, we need to compute C.
(1/3)ln(6-.5) - (1/3)ln(3-.5) = C
so, C is about .262819.
So, t = (1/3)ln(6-2.5) - (1/3)ln(3-2.5) - .262819 = 0.385818 minutes, which would be 23.149 seconds.