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#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

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

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

### differential_equations.pdf

drive.google.com

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