Why can I state $ \left| 3-y \right|=e^{ -4t }$ iff $3-y=e^{ -4t } $?

Question

I have this differential equation problem.

$$ \frac { dy }{ dt } =\quad 12-4y,\quad y\left( 0 \right) \quad =\quad 0 $$

Walking through my steps.

$$ \frac { dy }{ dt } =\quad -4\left( y-3 \right) \\ \frac { 1 }{ \left( y-3 \right) } dy\quad =\quad -4dt\\ \int { \frac { 1 }{ \left( y-3 \right) } dy } \quad =\quad -\int { 4dt } \\ \ln { \left( \left| 3-y \right| \right) } \quad =\quad -4t+c\\ { e }^{ \ln { \left( \left| 3-y \right| \right) } }=\quad { e }^{ -4t }+c\\ \left| 3-y \right| \quad =\quad { e }^{ -4t }+c $$

Now my teacher has stated that we can assume the LHS is always going to be positive. She attempted to explain this to me but I was unable to understand the proof/logic behind stating the following two steps $$ \left| 3-y \right| \quad =\quad { e }^{ -4t } +c\\ \quad 3-y\quad =\quad { e }^{ -4t }+c $$ How can I show that $$ { e }^{ -4t } $$ is always positive?

Answer 1

Given $\dfrac{dy}{dt}=12-4y=4(3-y),\quad y(0)=0$.

$$\int\dfrac{1}{3-y}\,dy=\int4\,dt$$

$$-\ln\vert 3-y\vert=4t-c_0$$

$$ \ln\vert 3-y\vert=c_0-4t $$

$$ \vert3-y\vert=e^{c_0}e^{-4t}=c_1e^{-4t} $$

$$3-y = \pm c_1{ e }^{ -4t }=ce^{-4t}$$

$$ y=3-ce^{-4t}$$ is the solution

$y(0)=3-c\cdot1=0$ so $c=3$. Therefore $y(t)=3-3e^{-4t}$.

Answer 2

When you integrate $-4dt$ you should get \begin{equation} -\ln(|3-y|) = -4t + A_1 \end{equation} where $C$ is an unknown constant. So your solution should be $|3 - y| = A_2e^{-4t}$ where $A_2 = e^{A_1}$. Therefore $3 - y = \pm A_2 e^{-4t}$ and we can just absorb $\pm$ into the constant $A_2$ and write $y = 3 + Ae^{-4t}$. The conclusion is you should get more than one solution for your ODE. Why is that? Because an ODE doesn't give a unique solution unless you have an initial condition $y(0) = $ something. Once an initial condition is known you can solve for $A$.

EDIT: Since $y(0) = 0$ we should have $A = -3$ ie. $y = 3 - 3e^{-4t}$.