Answer
3. (a) $y=ce^{-at} + b/a$.
(b) [In place of a sketch, here, we calculate the values of $a$, $b$ and $c$ for three different y-intercepts: $y(0)=1$, $y(0)=0$ and $y(0)=-1$, and write the actual equations for the corresponding values for $a$, $b$ and $c$ from which sketches could be made. Thus,
for $y(0)=1$, $a=-2$, $b=1$, $c=3/2$ and so the corresponding equation is $y=\frac{3}{2} e^{2t}-\frac{1}{2}$; for $y(0)=0$, $a=-1$, $b=1$ and $c=1$, giving the equation $y\equiv 0$; and for $y(0)=-1$, $a=2$, $b=1$ and $c=-3/2$, yielding the equation, $y=-\frac{3}{2}e^{-2t}+\frac{1}{2}$.]
(c) (i) Equilibrium is lower and is approached more rapidly.
(ii) Equilibrium is higher.
(iii) Equilibrium remains the same and is approached more rapidly,
Work Step by Step
3. Given the differential equation
$dy/dt = -ay+b$,
where both $a$ and $b$ are positive numbers,
(a) Solve the differential equation.
(b) Sketch the solution for several different initial conditions.
(c) Describe how the solutions change under each of the following conditions,
(i) $a$ increases.
(ii) $b$ increases.
(ii) Both $a$ and $b$ increase, but the ratio, $b/a$ remains the same.
(a) $dy/dt = -ay + b$,
$\frac{dy/dt}{y- b/a} = -a$,
$\ln|y-b/a| = -at + C$,
$|y-b/a| = e^{-at}\cdot e^{C}$ $=ce^{-at}$, where $c=e^{C}$,
$y=ce^{-at} + b/a$.
(b) [Rather than producing sketches at this point. we give particular equations for three different initial values, which will enable corresponding sketches (or displays on a graphing calculator).]
The equation $y= ce^{-at} + b/a$, evaluated for $t=0$, is
$y(0) = c + b/a$.
So, for $y(0) = 1$, let $c=3/2$; we then are forced to have $b/a = -1/2$, and for $b=1$, $a=-2$. We then get $y= \frac{3}{2} e^{2t} - \frac{1}{2}$.
For $y(0) =0$, $c=-b/a$, so letting $c=1$, $b/a =-1$ and with $b=1$, $a=-1$. Our particular equation, in this case, is the constant, $y=0$.
For $y(0)=-1$, $c+b/a = -1$, and with $c=-3/2$, $b/a = 1/2$ and with $b=1$, $a=2$, which then gives us the equation, $y=-\frac{3}{2}e^{-2t} + \frac{1}{2}$.
(c) (i) For our equation $y= ce^{-at} + b/a$, with $a$ increasing, we find by examining each of the two terms on the r,h,s,, we see that the equilibrium becomes lower (inferred from the second term) but is approached more rapidly (inferred from the first term on the r.h.s.)
(ii) With $b$ increasing, we infer from the second term of the r.h.s. that the equilibrium (where the graph of the solution is a horizontal) increases.
(iii) With both $a$ and $b$ increasing, but the ratio $b/a$ remaining the same, we see from the second term of the r.h.s. that the equilibrium remains the same but, from the first term on the r.h.s., we infer that it is approached more rapidly.
