GIVEN
[MATH]\dfrac{dx}{dt} = x'(t) = k(20 - x\}, \ x(0) = 0, \text { and} x'(0) = 1.[/MATH]
[MATH]\therefore 1 = k\{20 - x(0)\} \implies 20k = 1 \implies k = 0.05\implies\\
x' = 0.05(20 - x) \implies \dfrac{dx}{dt} = \dfrac{20 - x}{20}.[/MATH]Straightforward. We have an obviously separable equation.
[MATH]\int \dfrac{dx}{20 - x} = \int \dfrac{dt}{20} = 0.05 \int dt \implies\\
- \int \dfrac{-dx}{20 - x} = 0.05t \int dt \implies \\
- ln\{|c_1(20 - x)|\}= 0.05t \implies |c_1(20 - x)| = e^{-0.05t}.[/MATH]Can x be negative? No, that would make no sense in terms of the physical situation being described. But x can be zero. So we need to look at the case 0 < x < 20.
Can x ever = 20? No, because then the LHS would be zero so that t would have to be inifinite to make the RHS zero also. In other words, x starts at 0 but can never get to 20. That is what Dr. Peterson meant when he said there could be no "cross over." We can ignore x > 20.
[MATH]\therefore 0 \le x < 20 \implies c_1(20 - x) = e^{-0.05t} \implies \\
c_1(20 - 0) = e^{-0.05(0)} \implies 20c_1 = 1 \implies c_1 = 0.05 \implies \\
0.05(20 - x) = e^{-0.05t} \implies x = 20 - 20e^{-0.05t} \implies
[/MATH]Now let's check.
[MATH]\therefore \dfrac{dx}{dt} = 0 - 20e^{-0.05t}(-0.05) = e^{-0.05t} = 0.05(20 - x) \ \checkmark.[/MATH]
