Problem: Consider the initial value problem
\[
\frac{\mathrm{d}y}{\mathrm{d}x} + \alpha y = 0,~ y(0) = 1,~ \alpha \in \mathbb{R} .
\]
Find the value of $\alpha \in \mathbb{R} $ such that $y(1) = 2$.
Solution: We have
\begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x} + \alpha y = 0 & \implies \frac{\mathrm{d}y}{y} + \alpha \mathrm{d} x = 0 \\
& \implies \ln y + \alpha x = \ln c \\
& \implies y = c e^{- \alpha x} .
\end{align*}
Now we will use the initial condition to determine the value of the constant $c$.
\begin{align*}
y(0) = 1 & \implies c = 1.
\end{align*}
Therefore, $y(x) = e^{-\alpha x}$. Now given that
\begin{align*}
y(1) = 2 & \implies e^{-\alpha } = 2 \implies \alpha = -\ln 2.
\end{align*}