3.3.2.5 Example 5 \(y^{\prime }=F\left ( x^{2}+5y\right ) \)

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

\[ y^{\prime }=F\left ( x^{2}+5y\right ) \]

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

\begin{align*} \frac {u^{\prime }-2x}{5} & =F\left ( u\right ) \\ u^{\prime } & =5F\left ( u\right ) +2x \end{align*}

Which is no longer separable. Lets see what happens if \(y\) was not linear. Let the ode be

\[ y^{\prime }=F\left ( x+y^{2}\right ) \]

Let \(u=x+y^{2}\) then \(\frac {du}{dx}=1+2yy^{\prime }\). Hence \(y^{\prime }=\frac {u^{\prime }-1}{2y}\) and the ode becomes

\begin{align*} \frac {u^{\prime }-1}{2y} & =F\left ( u\right ) \\ u^{\prime } & =2yF\left ( u\right ) +1 \end{align*}

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 \(x,y\)