3.5.5.1 Example 1

$$\begin{array}{rl}{\left(y^{\prime }\right)}^4+f\left(x\right){(y-a)}^3{(y-b)}^3{(y-c)}^2& =0\\ {\left(y^{\prime }\right)}^4& =-f\left(x\right){(y-a)}^3{(y-b)}^3{(y-c)}^2\\ \frac{{\left(y^{\prime }\right)}^4}{{(y-a)}^3{(y-b)}^3{(y-c)}^2}& =-f\left(x\right)\\ {\left(\frac{y^{\prime }}{{\left({(y-a)}^3{(y-b)}^3{(y-c)}^2\right)}^{\frac{1}{4}}}\right)}^4& =-f\left(x\right)\\ \frac{y^{\prime }}{{\left({(y-a)}^3{(y-b)}^3{(y-c)}^2\right)}^{\frac{1}{4}}}& ={(-f\left(x\right))}^{\frac{1}{4}}\\ \frac{y^{\prime }}{{\left((y-a)(y-b){(y-c)}^{\frac{2}{3}}\right)}^{\frac{3}{4}}}& ={(-f\left(x\right))}^{\frac{1}{4}}\\ \frac{dy}{{\left((y-a)(y-b){(y-c)}^{\frac{2}{3}}\right)}^{\frac{3}{4}}}& ={(-f\left(x\right))}^{\frac{1}{4}}dx\\ {\int }^{y\left(x\right)}\frac{1}{{\left((z-a)(z-b){(z-c)}^{\frac{2}{3}}\right)}^{\frac{3}{4}}}dz& ={\int }^x{(-f\left(\tau \right))}^{\frac{1}{4}}d\tau +c_1\end{array}$$