5.5.1.2 Example 2

$${\left(y^{\prime }\right)}^{\frac{3}{2}}=2y+3x+9$$

Let $u=2y+3x+9$ then $u^{\prime }=2y^{\prime }+3$ then $y^{\prime }=\frac{u^{\prime }-3}{2}$and the ode becomes

$$\begin{array}{rl}{\left(\frac{u^{\prime }-3}{2}\right)}^{\frac{3}{2}}& =u\\ {\left({\left(\frac{u^{\prime }-3}{2}\right)}^{\frac{1}{2}}\right)}^3& =u\end{array}$$

Let ${\left(\frac{u^{\prime }-3}{2}\right)}^{\frac{1}{2}}=Y$ then

$$\begin{array}{rl}{Y}^3& =u\\ Y& =\{\begin{array}{c}{u}^{\frac{1}{3}}\\ u^{\frac{1}{3}}{(-\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{\frac{1}{3}}\\ u^{\frac{1}{3}}{(-\frac{1}{2}-\frac{i\sqrt{3}}{2})}^{\frac{1}{3}}\end{array}\end{array}$$

Hence

$$\begin{array}{rl}{\left(\frac{u^{\prime }-3}{2}\right)}^{\frac{1}{2}}& =\{\begin{array}{c}{u}^{\frac{1}{3}}\\ u^{\frac{1}{3}}{(-\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{\frac{1}{3}}\\ u^{\frac{1}{3}}{(-\frac{1}{2}-\frac{i\sqrt{3}}{2})}^{\frac{1}{3}}\end{array}\\ \left(\frac{u^{\prime }-3}{2}\right)& =\{\begin{array}{c}{u}^{\frac{2}{3}}\\ u^{\frac{2}{3}}{(-\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}\\ u^{\frac{2}{3}}{(-\frac{1}{2}-\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}\end{array}\\ u^{\prime }& =\{\begin{array}{c}2u^{\frac{2}{3}}+3\\ 2u^{\frac{2}{3}}{(-\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}+3\\ 2u^{\frac{2}{3}}{(-\frac{1}{2}-\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}+3\end{array}\end{array}$$

Each is solved as separable.

$$\{\begin{array}{c}\int \frac{du}{2u^{\frac{2}{3}}+3}=\int dx\\ \int \frac{du}{2u^{\frac{2}{3}}{(-\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}+3}=\int dx\\ \int \frac{du}{2u^{\frac{2}{3}}{(-\frac{1}{2}-\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}+3}=\int dx\end{array}$$

Hence the three solutions are

$$\{\begin{array}{c}{\int }^{2y\left(x\right)+3x+9}\frac{dz}{2z^{\frac{2}{3}}+3}=x+c_1\\ {\int }^{2y\left(x\right)+3x+9}\frac{dz}{2z^{\frac{2}{3}}{(-\frac{1}{2}+\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}+3}=x+c_1\\ {\int }^{2y\left(x\right)+3x+9}\frac{dz}{2z^{\frac{2}{3}}{(-\frac{1}{2}-\frac{i\sqrt{3}}{2})}^{\frac{2}{3}}+3}=x+c_1\end{array}$$