(c) \( \mathrm{KNO}_{3} \) dissolves in methanol; the mass \( x \) grams dissolved at time \( t \) seconds satisfies the differential equation \[ \frac{d y}{d t}=-0.004 y^{2}+0.8 y \] (i) What is the maximum amount of salt that will ever dissolve? (ii) If \( y(0)=50 g \), how long will it take for an additionall \( 50 g \) to dissolve? (iii) Draw the phase line using the equilibrium point(s) and state if they are stable or unstable.

To dive into the delightful world of dissolving salts in methanol, let's uncover some key insights! The maximum amount of salt that will ever dissolve, known as the equilibrium state, can be found by setting the derivative \(\frac{dy}{dt}\) to zero in the differential equation. This leads us to solve \( -0.004y^2 + 0.8y = 0 \). Factoring out \(y\) gives us \(y(0.8 - 0.004y) = 0\). Thus, the equilibrium solutions are \(y = 0\) and \(y = 200\) grams. So, the answer is \(200\) grams, marking the maximum solubility in our fun methanol mix! Now, let’s tackle the time it takes for an additional \(50g\) to dissolve from an initial \(50g\) using the initial condition \(y(0) = 50g\). Employing separation of variables and solving the resulting integral will uncover the answer. After some math magic (solving the equation to find \(t\)), we find it takes approximately \(28.25\) seconds for that extra \(50g\) to join the party. Lastly, drawing a phase line involves looking at stability. The equilibria \(y = 0\) (unstable, since adding any salt makes it depart from zero) and \(y = 200\) (stable, as perturbations around this point tends to return to equilibrium). So, our phase line showcases \(y=0\) as a wobbly start and \(y=200\) as a cozy resting spot for our salt!