This example shows why this method works only when the argument of \(F\left ( \cdot \right ) \) is linear in \(x\) and \(y\). Assuming the ode given is
Which is not linear in \(x,y\). As before, let \(u=x^{2}+5y\) then \(\frac {du}{dx}=2x+5y^{\prime }\). Hence \(y^{\prime }=\frac {u^{\prime }-2x}{5}\) and the ode becomes
Which is no longer separable. Lets see what happens if \(y\) was not linear. Let the ode be
Let \(u=x+y^{2}\) then \(\frac {du}{dx}=1+2yy^{\prime }\). Hence \(y^{\prime }=\frac {u^{\prime }-1}{2y}\) and the ode becomes
We see that the term \(y\) did not vanish and this can not work. This shows that for this method to work, the argument of the function \(F\) must be linear in \(x,y\)