2.6.1.3 Example 3 \(y^{\prime }=f\left ( y\right ) g\left ( x\right ) \)

Solve

\[ y^{\prime }=f\left ( y\right ) g\left ( x\right ) \]

Such that \(f\left ( y\right ) g\left ( x\right ) \) is continuous everywhere and \(f_{y}g\) is also. Hence it is guaranteed that solution exist and unique. Let initial conditions be such that \(f\left ( y_{0}\right ) =0\). For example, if \(f\left ( y\right ) =y\) and \(y\left ( 0\right ) =0\). In this case, we can not separate using

\[ \frac {dy}{f\left ( y\right ) }=g\left ( x\right ) \qquad f\left ( y\right ) \neq 0 \]

Since \(f\left ( y\right ) =0\) at I.C. So we use the short cut method. Substituting IC into the ode gives

\begin{align*} y^{\prime } & =0\\ y & =c \end{align*}

But since the solution is unique, then \(C_{1}=0\) since \(y=0\) is given and only one solution \(y\left ( x\right ) \) can exist. Hence this is the solution.

\[ y=0 \]

So the bottom line is this: Given a first order ode \(y^{\prime }=f\left ( y\right ) g\left ( x\right ) \) where the solution exist and unique and \(f\left ( y\right ) =0\) at IC, then the solution is always

\[ y=0 \]

Lets look at another special case ode.