Example 5

\begin{align*} y^{\prime } & =y\left ( y-1\right ) \left ( y-3\right ) \\ y\left ( 0\right ) & =4 \end{align*}

A solution exist an is unique. Integrating gives

\begin{align} \int \frac {dy}{y\left ( y-1\right ) \left ( y-3\right ) } & =\int dx\qquad y\left ( y-1\right ) \left ( y-3\right ) \neq 0\nonumber \\ \frac {1}{3}\ln y+\frac {1}{6}\ln \left ( y-3\right ) -\frac {1}{2}\ln \left ( y-1\right ) & =x+c_{1} \tag {1}\end{align}

Applying initial conditions gives

\begin{align*} \frac {1}{3}\ln 4+\frac {1}{6}\ln \left ( 1\right ) -\frac {1}{2}\ln \left ( 3\right ) & =c_{1}\\ \frac {1}{3}\ln 4-\frac {1}{2}\ln \left ( 3\right ) & =c_{1}\end{align*}

Hence the solution from (1) is

\[ \frac {1}{3}\ln y+\frac {1}{6}\ln \left ( y-3\right ) -\frac {1}{2}\ln \left ( y-1\right ) =x+\frac {1}{3}\ln 4-\frac {1}{2}\ln \left ( 3\right ) \]

Lets see what happens if we convert to exponential first. Applying exponential to both sides of (1) gives

\begin{align} \exp \left ( \ln y^{\frac {1}{3}}+\ln \left ( y-3\right ) ^{\frac {1}{6}}+\ln \left ( y-1\right ) ^{\frac {-1}{2}}\right ) & =c_{2}e^{x}\nonumber \\ y^{\frac {1}{3}}\left ( y-3\right ) ^{\frac {1}{6}}\left ( \frac {1}{\sqrt {y-1}}\right ) & =c_{2}e^{x}\nonumber \\ \frac {y^{\frac {1}{3}}\left ( y-3\right ) ^{\frac {1}{6}}}{\sqrt {y-1}} & =c_{2}e^{x} \tag {2}\end{align}

At IC

\begin{align*} \frac {4^{\frac {1}{3}}\left ( 4-3\right ) ^{\frac {1}{6}}}{\sqrt {4-1}} & =c_{2}\\ \frac {4^{\frac {1}{3}}}{\sqrt {3}} & =c_{2}\end{align*}

Hence the solution from (2) is

\[ \frac {y^{\frac {1}{3}}\left ( y-3\right ) ^{\frac {1}{6}}}{\sqrt {y-1}}=\frac {4^{\frac {1}{3}}}{\sqrt {3}}e^{x}\]

And this is also correct. I prefer to convert to exponential when the solution has the form   \(f\left ( y\right ) =cg\left ( x\right ) \) where \(f\left ( y\right ) \) is made up of all \(\ln \) as functions of \(y\). This makes finding constant of integration easier in all cases.