I am given $$y'=0.05y-800$$ I am asked to:
(a) Find all constant solutions of the differential equation.
(b) Suppose $y = M$ is your constant solution from (a). Plot two solutions of the differential equation where one satisfies $ 0 < y(0) <M$ and the other satisfying $y(0) > M$. Use this sketch for parts (c), (d) and (e).
Here is what I did. For part a) I equated $y'$ to $0$ since I needed the constant solution for $y'$ and i got $y=16000$. For part b, I'm truly confused as to why there are two solutions for this one. I have the key answer for these problems but at some point I don't how they got the answer and my exam is fast approaching. I need to know how. Please help. Thanks!
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$\begingroup$Your solution to (a) is correct.
For (b), there are actually infinitely many solutions $\Bbb R \to \Bbb R$, each of which corresponds to a different value of $y(0)$. (NB this equation is separable, and one can solve it explicitly by integrating.)
$\endgroup$ 2 $\begingroup$$$ \frac{dy}{dx}=0.05y-800 $$ $$ \frac{dy}{0.05 y - 800} = dx $$ $$ \frac{20\,dy}{y-16000} = dx $$ $$ 20\log|y-16000| = x + C_1 $$ $$ \log|y-16000| = \frac x {20} + C_2\qquad(\text{i.e.}~C_2=C_1/20) $$ $$ |y-16000| = e^{x/20}\cdot C_3\qquad(\text{i.e.}~C_3=e^{C_2}\text{ is a positive constant.}) $$ $$ y-16000 = e^{x/20}\cdot C_4\quad\text{and $C_4$ need not be positive.} $$ $$ y = 16000 + C e^{x/20}. $$
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