Solving \(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {1}{n}}\)

For \(n\) integer \(\neq 1\) which can be negative or positive, the ode is

\begin{equation} \frac {dy}{dx}=\left ( a+bx+cy\right ) ^{\frac {1}{n}} \tag {1}\end{equation}

Let \(z=a+bx+cy\) then

\begin{align*} \frac {dz}{dx} & =b+c\frac {dy}{dx}\\ \frac {dy}{dx} & =\left ( \frac {dz}{dx}-b\right ) \frac {1}{c}\end{align*}

Hence (1) becomes

\begin{align} \left ( \frac {dz}{dx}-b\right ) \frac {1}{c} & =z^{\frac {1}{n}}\nonumber \\ \frac {dz}{dx} & =cz^{\frac {1}{n}}+b\nonumber \\ \int \frac {dz}{cz^{\frac {1}{n}}+b} & =\int dx \tag {2}\end{align}

If the left side is integrable, then the solution to (1) can be found. For \(n\) integer it is possible to find antiderivative. For example for \(n=2\) then (2) becomes

\[ \frac {2}{c}\sqrt {z}-\frac {2b\ln \left ( b+c\sqrt {z}\right ) }{c^{2}}=x+C_{1}\]

Replacing back \(z=a+bx+cy\) the above becomes

\begin{equation} \frac {2}{c}\sqrt {a+bx+cy}-\frac {2b\ln \left ( b+c\sqrt {a+bx+cy}\right ) }{c^{2}}=x+C_{1} \tag {3}\end{equation}

Which is the implicit solution to (1).

To show that the above does not work if we had \(xy\) term, lets give an example. Let \(y^{\prime }=\left ( a+xy\right ) ^{\frac {1}{2}}\), then following the above, let \(z=a+xy\) and \(\frac {dz}{dx}=y+xy^{\prime }\) or \(y^{\prime }=\frac {\frac {dz}{dx}-y}{x}\). Hence \(z^{\frac {1}{2}}=\frac {\frac {dz}{dx}-y}{x}\) or \(xz^{\frac {1}{2}}+y=\frac {dz}{dx}\) and this is not separable. (it is Chini ode, where is very hard to solve).

for \(n=2\). Using \(a=1,b=1,c=1\) Eq. (1) becomes

\[ \frac {dy}{dx}=\left ( 1+x+y\right ) ^{\frac {1}{2}}\]

And (3) becomes

\begin{equation} 2\sqrt {1+x+y}-2\ln \left ( 1+\sqrt {1+x+y}\right ) =x+C_{1} \tag {4}\end{equation}

And for \(n=3\) Eq. (2) becomes

\[ \frac {3\left ( -2b+cz^{\frac {1}{3}}\right ) }{2c^{2}}z^{\frac {1}{3}}+\frac {3b^{2}\ln \left ( b+cz^{\frac {1}{3}}\right ) }{c^{3}}=x+C_{1}\]

Replacing back \(z=a+bx+cy\) the above becomes

\begin{equation} \frac {3\left ( -2b+c\left ( a+bx+cy\right ) ^{\frac {1}{3}}\right ) }{2c^{2}}z^{\frac {1}{3}}+\frac {3b^{2}\ln \left ( b+c\left ( a+bx+cy\right ) ^{\frac {1}{3}}\right ) }{c^{3}}=x+C_{1} \tag {5}\end{equation}

Which is the implicit solution to (1) for \(n=3\). Using \(a=1,b=1,c=1\) then (1) becomes

\[ \frac {dy}{dx}=\left ( 1+x+y\right ) ^{\frac {1}{3}}\]

And its solution (5) becomes

\[ \frac {3}{2}\left ( -2+\left ( 1+x+y\right ) ^{\frac {1}{3}}\right ) \left ( 1+x+y\right ) ^{\frac {1}{3}}+3\ln \left ( 1+\left ( 1+x+y\right ) ^{\frac {1}{3}}\right ) =x+C_{1}\]

And so on for higher values of \(n\). This also works negative values of \(n\). For example, for \(n=-2\) then (1) becomes

\[ \frac {dy}{dx}=\left ( a+bx+cy\right ) ^{\frac {-1}{2}}\]

And the integral equation (2) now becomes

\[ \int \frac {dz}{cz^{\frac {-1}{n}}+b}=\int dx \]

Which for \(n=2\) gives

\[ \frac {1}{b^{3}}\left ( -2bc\sqrt {z}+b^{2}z+2c^{2}\ln \left ( c+b\sqrt {z}\right ) \right ) =x+C_{1}\]

Replacing back \(z=a+bx+cy\) the above becomes

\[ \frac {1}{b^{3}}\left ( -2bc\sqrt {a+bx+cy}+b^{2}\left ( a+bx+cy\right ) +2c^{2}\ln \left ( c+b\sqrt {a+bx+cy}\right ) \right ) =x+C_{1}\]

For \(a=1,b=1,c=1\) the above becomes

\[ \left ( -2\sqrt {1+x+y}+\left ( 1+x+y\right ) +2\ln \left ( 1+\sqrt {1+x+y}\right ) \right ) =x+C_{1}\]

And so on.