Initial Values Problem

Problem 190: Solve the following differential equation given the initial condition:

(1) $\begin{equation*} \frac{dy}{dx} = \frac{1}{x}, \quad y(\pi) = -3. \end{equation*}$

Solution: In order to solve this problem, we need to find integrate both sides. That is,

(2) $\begin{align*} \frac{dy}{dx} & = \frac{1}{x} \\ dy & = \frac{1}{x}dx \\ \int dy & = \int \frac{1}{x} dx \\ y(x) & = \ln(x) + C. \end{align*}$

Using our initial condition, we can find $C$ . Thus

(3) $\begin{align*} y(x) & = \ln(x) + C \\ y(\pi) & = \ln(\pi) + C = -3 \\ C & = -\ln(\pi) - 3. \end{align*}$

Hence,

(4) $\begin{align*} y(x) & = \ln(x) + C \\ & = \ln(x) - \ln(\pi) - 3 \\ & = \ln\left(\frac{x}{\pi} \right) -3. \end{align*}$