2.1.63.1 Solved using first_order_ode_homog_type_C

Entering first order ode homog type C solver

\[\begin {aligned} y^{\prime }&=\left (a +b x +y\right )^{4} \end {aligned}\]

Let

\begin{align*} z = a +b x +y\tag {1} \end{align*}

Then

\begin{align*} z^{\prime }\left (x \right )&=b +y^{\prime } \end{align*}

Therefore

\begin{align*} y^{\prime }&=z^{\prime }\left (x \right )-b \end{align*}

Hence the given ode can now be written as

\begin{align*} z^{\prime }\left (x \right )-b&=z^{4} \end{align*}

This is separable first order ode. Integrating

\begin{gather*} \begin {aligned} \int d x&=\int \frac {1}{z^{4}+b}d z\\ x +c_1&=\frac {\sqrt {2}\, \left (\ln \left (\frac {z^{2}+b^{{1}/{4}} z \sqrt {2}+\sqrt {b}}{z^{2}-b^{{1}/{4}} z \sqrt {2}+\sqrt {b}}\right )+2 \arctan \left (\frac {\sqrt {2}\, z}{b^{{1}/{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, z}{b^{{1}/{4}}}-1\right )\right )}{8 b^{{3}/{4}}} \end {aligned} \end{gather*}

Replacing \(z\) back by its value from (1) then the above gives the solution as

\begin{align*} \frac {\sqrt {2}\, \left (\ln \left (\frac {\left (a +b x +y\right )^{2}+b^{{1}/{4}} \left (a +b x +y\right ) \sqrt {2}+\sqrt {b}}{\left (a +b x +y\right )^{2}-b^{{1}/{4}} \left (a +b x +y\right ) \sqrt {2}+\sqrt {b}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +b x +y\right )}{b^{{1}/{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +b x +y\right )}{b^{{1}/{4}}}-1\right )\right )}{8 b^{{3}/{4}}} = x +c_1 \end{align*}

Simplifying the above gives

\[ \frac {\sqrt {2}\, \left (\ln \left (\frac {\sqrt {2}\, b^{{5}/{4}} x +\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+2 \left (b x +a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}{-\sqrt {2}\, b^{{5}/{4}} x -\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+2 \left (b x +a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +b x +y\right )}{b^{{1}/{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +b x +y\right )}{b^{{1}/{4}}}-1\right )\right )}{8 b^{{3}/{4}}} = x +c_1 \]

Summary of solutions found

\[ \frac {\sqrt {2}\, \left (\ln \left (\frac {\sqrt {2}\, b^{{5}/{4}} x +\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+2 \left (b x +a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}{-\sqrt {2}\, b^{{5}/{4}} x -\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+2 \left (b x +a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +b x +y\right )}{b^{{1}/{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +b x +y\right )}{b^{{1}/{4}}}-1\right )\right )}{8 b^{{3}/{4}}} = x +c_1 \]