ode internal name "first_order_nonlinear_p_but_linear_in_x_y"
For the special case of \(\left ( y^{\prime }\right ) ^{\frac {n}{m}}=F\left ( x,y\right ) \) where RHS is linear in both \(x\) and \(y\), i.e. \(F\left ( x,y\right ) =ax+by+c\) then a short cut method is described below using transformation \(u=ax+by+c\). This makes it separable in \(u\). This will not work if there is nonlinear \(x\) term, such as \(\left ( y^{\prime }\right ) ^{\frac {n}{m}}=by+x^{2}\) or nonlinear term in \(y\) such as \(\left ( y^{\prime }\right ) ^{\frac {n}{m}}=y^{2}+x\).
Taking derivatives gives \(u^{\prime }=a+by^{\prime }\) or \(y^{\prime }=\frac {u^{\prime }-a}{b}\) and the ode becomes\begin {align*} \left ( \frac {u^{\prime }-a}{b}\right ) ^{\frac {n}{m}} & =u\\ \left ( \frac {u^{\prime }-a}{b}\right ) ^{n} & =u^{m} \end {align*}
Here we need to find roots of unity for \(n\). For example, for \(n=2\) we have\[ \frac {u^{\prime }-a}{b}=\left \{ \begin {array} [c]{c}\left ( u\right ) ^{\frac {m}{2}}\\ -\left ( u\right ) ^{\frac {m}{2}}\end {array} \right . \] And for \(n=3\)\[ \frac {u^{\prime }-a}{b}=\left \{ \begin {array} [c]{c}\left ( u\right ) ^{\frac {m}{3}}\\ -\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {m}{3}}\\ \left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {m}{3}}\end {array} \right . \] And so on. From now on, this is solved as separable. For negative integer values \(n\), we just replaced \(n\) by \(-n\) in the above. For example, for \(n=3\)\[ \frac {u^{\prime }-a}{b}=\left \{ \begin {array} [c]{c}\left ( u\right ) ^{\frac {m}{-3}}\\ -\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {m}{-3}}\\ \left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {m}{-3}}\end {array} \right . \] For symbolic values of \(n\) we can just leave the integral as is. For example for \(\left ( y^{\prime }\right ) ^{r}=ax+by\) we obtain\begin {align*} \left ( \frac {u^{\prime }-a}{b}\right ) ^{r} & =u\\ \frac {u^{\prime }-a}{b} & =u^{\frac {1}{r}}\\ u^{\prime } & =bu^{\frac {1}{r}}+a\\ \int \frac {du}{bu^{\frac {1}{r}}+a} & =\int dx+c_{1}\\ \int ^{ax+by\left ( x\right ) }\frac {dz}{bz^{\frac {1}{r}}+a} & =x+c_{1} \end {align*}