- #1

karush

Gold Member

MHB

- 3,269

- 5

$\textit{given $y''+y'-2y=0\quad y(0)=1 \quad y'(0)=1$ }$

\begin{align*}\displaystyle

\textit{if } r&=e^{t} \textit{then:}\\

y''+y'-2y&=r^2+r-2=(r+2)(r-1)=0\\

&=c_1^{2t}+c_2^{-t}\\

y&=\color{red} {e^t; \quad y \to \infty \,as \, t \to \infty }

\end{align*}ok the bk answer is in red

I just don't get this

why do we need $$y(0)=1 \quad y'(0)=1$$

the graph of $y=e^t$

\begin{align*}\displaystyle

\textit{if } r&=e^{t} \textit{then:}\\

y''+y'-2y&=r^2+r-2=(r+2)(r-1)=0\\

&=c_1^{2t}+c_2^{-t}\\

y&=\color{red} {e^t; \quad y \to \infty \,as \, t \to \infty }

\end{align*}ok the bk answer is in red

I just don't get this

why do we need $$y(0)=1 \quad y'(0)=1$$

the graph of $y=e^t$

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