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.