3.3.23.3 Example 3

\[ y^{\prime }=ay^{5}+bx^{\left ( -\frac {5}{4}\right ) }\]

This is Kamke 1.52. First we find the Chini invariant. It should come out as constant. We see that \(f=a,g=0,h=bx^{-\frac {5}{4}},n=5\), hence

\begin{align*} \Delta & =f^{-n-1}h^{-2n+1}\left ( fh^{\prime }-f^{\prime }h-ngfh\right ) ^{n}n^{-n}\\ & =-\frac {1}{1024}\frac {1}{ab^{4}}\end{align*}

The above \(\Delta \) is also used in the solution below. It is a constant in this example, hence can be solved. Now we follow Kamke method to actually solve the ode. Now we need to find \(\alpha \). This can be found more easily from EQ (4)

\begin{equation} z^{\prime }-gz=\alpha h \tag {4}\end{equation}

Where \(z=\left ( \frac {h}{f}\right ) ^{\frac {1}{n}}=\left ( \frac {bx^{-\frac {5}{4}}}{a}\right ) ^{\frac {1}{5}}=\left ( \frac {b}{a}\right ) ^{\frac {1}{5}}x^{-\frac {1}{4}}\). Hence \(z^{\prime }=-\left ( \frac {b}{a}\right ) ^{\frac {1}{5}}\frac {1}{4}x^{-\frac {5}{4}}\). Therefore (4) becomes (given that \(g=0\))

\begin{align*} -\left ( \frac {b}{a}\right ) ^{\frac {1}{5}}\frac {1}{4}x^{-\frac {5}{4}} & =\alpha bx^{-\frac {5}{4}}\\ \alpha & =-\frac {1}{4a^{\frac {1}{5}}b^{\frac {4}{5}}}\end{align*}

\(\allowbreak \)Since \(\Delta \) is not zero, then solution is directly given as (from Kamke)

\begin{align*} \int ^{\alpha \left ( \frac {h}{f}\right ) ^{\frac {-1}{n}}y\left ( x\right ) }\frac {1}{\frac {u^{n}}{\Delta }-u+1}du-\int \alpha \left ( \frac {h}{f}\right ) ^{\frac {-1}{n}}hdx+c_{1} & =0\\ \int ^{-\frac {1}{4a^{\frac {1}{5}}b^{\frac {4}{5}}}\left ( \frac {bx^{-\frac {5}{4}}}{a}\right ) ^{-\frac {1}{5}}y\left ( x\right ) }\frac {1}{-1024ab^{4}u^{4}-u+1}du+\int -\frac {1}{4a^{\frac {1}{5}}b^{\frac {4}{5}}}\left ( \frac {bx^{-\frac {5}{4}}}{a}\right ) ^{-\frac {1}{5}}bx^{-\frac {5}{4}}dx+c_{1} & =0\\ \int ^{-\frac {x^{\frac {1}{4}}}{4b}y\left ( x\right ) }\frac {1}{-1024ab^{4}u^{4}-u+1}du+\int -\frac {1}{4x}dx+c_{1} & =0\\ \int ^{-\frac {x^{\frac {1}{4}}}{4b}y\left ( x\right ) }\frac {1}{-1024ab^{4}u^{4}-u+1}du-\frac {1}{4}\ln \left ( x\right ) +c_{1} & =0 \end{align*}

Note: In the above two examples \(\Delta \) was not zero. What to do if we obtain \(\Delta =0\) ? in this case, the solution becomes

\[ \int ^{\left ( \frac {h}{f}\right ) ^{\frac {-1}{n}}}\frac {1}{u^{n}+1}du-\int \left ( \frac {h}{f}\right ) ^{\frac {-1}{n}}hdx+c_{1}=0 \]