-b.3.1.9 Find the solution of y''+y'-2y=0\quad y(0)=1; y'(0)=1.

[karush]

$\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 $y=e^t$

 

[Opalg]

$\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 auxiliary equation $r^2 + r - 2 = 0$ factorises as $(r+2)(r-1)=0$, with solutions $r = -2$ and $r=1$. That tells you that the differential equation has solutions $y = e^{-2t}$ and $y = e^t$. The general solution is then $y = c_1e^{-2t} + c_2e^t$, where $c_1$ and $c_2$ are constants. To find the value of those constants, you need to use the conditions $y(0)=1$ and $y'(0)=1$. In fact, if you put $t=0$ in the equation $y = c_1e^{-2t} + c_2e^t$, it tells you that $1 = c_1 + c_2$. If you then differentiate the formula for $y$ and again put $t=0$, it will give you another equation for $c_1$ and $c_2$. You should find that $c_1=0$ and $c_2=1$, so that the solution becomes $y=e^t$.