MHB -b.1.1.4 directional field as t \to \infny

[karush]

Spoiler: problem source

from the spoiler its #4 hope I got instructions right

$y'=-1-2y$
$\begin{array}{ll}
rewrite &y'+2y=-1\\
exp &u(t)=\exp\ds\int 2 \, dx=e^{2t}\\
product &(e^{2t}y)'=-e^{2t}\\
&e^{2t}y=\ds\int -e^{2t} \, dt=\dfrac{-e^(2 t)}{2}+c\\
hence &y(t)=\dfrac{-e^{2t}}{2} + \dfrac{c}{e^{2t}}\\
t \to \infty &=-\dfrac{1}{2}+0\\
so &y \to \dfrac{1}{2}\textit{ as t} \to \infty
\end{array}$

ok this is an early problem but probably some comments to make it more understandable
hopefully correct
I was going to try tikz on this but probably 3 pages of code so its a desmos templatehttps://dl.orangedox.com/geAQogxCM0ZYQLaUju

 

[skeeter]

I get

$y = -\dfrac{1}{2} + Ce^{-2t}$

 

[HOI]

From $y'= -1- 2y$, you don't have to write it as $y'+ y= -1$.
You can rewrite it as $\frac{dy}{dx}= -1-2y$ and then separate as $\frac{dy}{2y+ 1}= -dx$..

Now integrate both sides- on the left, let $u= 2y+ 1$ so that $du= 2dy$ and $dy= \frac{1}{2}du$
$\frac{1}{2}\int \frac{du}{u}=\frac{1}{2}ln(u)+ c_1= \frac{1}{2} ln(2y+ 1)= ln(\sqrt{2y+ 1})$ and $-\int dx= -x+ c_2$.

Combining the two contants into "c", $ln(\sqrt{2y+ 1})= -x+c$. Taking the exponential of both sides $\sqrt{2y+1}= e^{-x+ c}= Ce^{-x}$ where $C= e^c$.

Square both sides: $2y+ 1= \left(Ce^{-x}\right)^2= C'e^{-2x}$ where $C'= C^2$.
$2y= C'e^{-2x}- 1$
$y= C''e^{-2x}- \frac{1}{2}$ where $C''= C'/2$

And that goes to $-\frac{1}{2}$ as x goes to infinity.

Hey, wasn't that fun!