-b.1.1.2 behavior of y'-2y=-3 as t goes to infinity

[karush]

determine the behavior of y as t →∞.
If this behavior depends on the initial value of y at t = 0,describe the dependency

\begin{array}{lll}
\textit{rewrite}
&y'-2y=-3\\ \\
u(t)
&=\exp\int -2 \, dx=e^{-2t}\\ \\
\textit{product}
&(e^{-2t}y)'=-3e^{-2t}\\ \\
\textit{integrate}
&e^{-2t}y=\int -3e^{-2t} \, dt =\dfrac{3e^{-2t}}{2}+c\\ \\
%e^{-2t}y&=\dfrac{3e^{-2t}}{2}+c\\ \\
\textit{isolate}
&y(t)=\dfrac{3}{2}+\dfrac{c}{e^{-2t}}\\ \\
t \to \infty&=\dfrac{3}{2}+0 =\dfrac{3}{2} \\ \\
\textit{so}
&y \textit{ diverges from } \dfrac{3}{2} \textit{ as t} \to \infty
\end{array}

ok think I got it ok
suggestions, typos, mother in law comments welcome

https://drive.google.com/file/d/17AneIqlG0aGlPXEQ8q8_kgCINV_ojeqx/view?usp=sharing

 

[Prove It]

I would have written $\displaystyle c\,\mathrm{e}^{2\,t}$ instead of $\displaystyle \frac{c}{\mathrm{e}^{-2\,t}} $.

As $\displaystyle t\to \infty, \,\,c\,\mathrm{e}^{2\,t} \to \infty $, so $\displaystyle y \to \infty $.

 

[karush]

yes better

 

[HOI]

Equivalently, the "linear differential equation with constant coefficients", $\frac{dy}{dt}- 2y= -3$ has "associated homogeneous part" $\frac{dy}{dx}- 2y= 0$. $\frac{dy}{dt}= 2y$, $\frac{dy}{y}= 2dt$.

Integrating, $ln|y|= 2t+ c$, $y= e^{2t+ c}= Ce^{2t}$ (C= e^c).

Since the non-homogeneus part of the equation is a constant, -3, we look for a solution to the entire equation of the form $y= A$, a constant. Then $\frac{dy}{dt}- 2y= 0- 2A= -3$ so $A= \frac{2}{3}$.

The general solutioin to the entire equation is $y(t)= Ce^{2t}+ \frac{2}{3}$.

The behavior "at infinity" is entirely determined by the exponential: $\lim_{t\to\infty} y(t)= \infty$.