3.5.6.1 Example 1

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

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

\begin{align*} \left ( \frac {u^{\prime }-3}{2}\right ) ^{3} & =u\\ \frac {u^{\prime }-3}{2} & =\left \{ \begin {array} [c]{c}\left ( u\right ) ^{\frac {1}{3}}\\ -\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {1}{3}}\\ \left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {1}{3}}\end {array} \right . \\ u^{\prime }-3 & =\left \{ \begin {array} [c]{c}2\left ( u\right ) ^{\frac {1}{3}}\\ -2\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {1}{3}}\\ 2\left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {1}{3}}\end {array} \right . \\ u^{\prime } & =\left \{ \begin {array} [c]{c}2\left ( u\right ) ^{\frac {1}{3}}+3\\ -2\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {1}{3}}+3\\ 2\left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {1}{3}}+3 \end {array} \right . \end{align*}

Each is now solved as separable.

\begin{align*} u^{\prime } & =2\left ( u\right ) ^{\frac {1}{3}}+3\\ \frac {du}{2\left ( u\right ) ^{\frac {1}{3}}+3} & =dx\\ \int \frac {du}{2\left ( u\right ) ^{\frac {1}{3}}+3} & =\int dx\\ \int \frac {du}{2\left ( u\right ) ^{\frac {1}{3}}+3} & =x+c_{1}\end{align*}

Hence

\[ \int ^{2y\left ( x\right ) +3x+9}\frac {dz}{2z^{\frac {1}{3}}+3}=x+c_{1}\]

For the second one \(u^{\prime }=-2\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {1}{3}}+3\) results in

\begin{align*} \frac {du}{-2\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {1}{3}}+3} & =dx\\ \int \frac {du}{-2\left ( -1\right ) ^{\frac {1}{3}}\left ( u\right ) ^{\frac {1}{3}}+3} & =\int dx\\ \int ^{2y\left ( x\right ) +3x+9}\frac {dz}{-2\left ( -1\right ) ^{\frac {1}{3}}\left ( z\right ) ^{\frac {1}{3}}+3} & =x+c_{1}\end{align*}

And for the third ode \(u^{\prime }=2\left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {1}{3}}+3\)

\begin{align*} \frac {du}{2\left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {1}{3}}+3} & =dx\\ \int \frac {du}{2\left ( -1\right ) ^{\frac {2}{3}}\left ( u\right ) ^{\frac {1}{3}}+3} & =\int dx\\ \int ^{2y\left ( x\right ) +3x+9}\frac {dz}{2\left ( -1\right ) ^{\frac {2}{3}}\left ( z\right ) ^{\frac {1}{3}}+3} & =x+c_{1}\end{align*}

Hence the three solutions are

\[ \left \{ \begin {array} [c]{c}\int ^{2y\left ( x\right ) +3x+9}\frac {dz}{2z^{\frac {1}{3}}+3}=x+c_{1}\\ \int ^{2y\left ( x\right ) +3x+9}\frac {dz}{-2\left ( -1\right ) ^{\frac {1}{3}}\left ( z\right ) ^{\frac {1}{3}}+3}=x+c_{1}\\ \int ^{2y\left ( x\right ) +3x+9}\frac {dz}{2\left ( -1\right ) ^{\frac {2}{3}}\left ( z\right ) ^{\frac {1}{3}}+3}=x+c_{1}\end {array} \right . \]