1.64 problem 64

Internal problem ID [8028]
Book : First order enumerated odes
Section : section 1
Problem number : 64
Date solved : Monday, October 21, 2024 at 04:43:56 PM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

Solve

\begin{align*} y^{\prime }&=\left (\pi +x +7 y\right )^{{7}/{2}} \end{align*}

1.64.1 Solved as first order homogeneous class C ode

Time used: 0.830 (sec)

Let

\begin{align*} z = \pi +x +7 y\tag {1} \end{align*}

Then

\begin{align*} z^{\prime }\left (x \right )&=1+7 y^{\prime } \end{align*}

Therefore

\begin{align*} y^{\prime }&=\frac {z^{\prime }\left (x \right )}{7}-\frac {1}{7} \end{align*}

Hence the given ode can now be written as

\begin{align*} \frac {z^{\prime }\left (x \right )}{7}-\frac {1}{7}&=z^{{7}/{2}} \end{align*}

This is separable first order ode. Integrating

\begin{align*} \int d x&=\int \frac {1}{7 z^{{7}/{2}}+1}d z \\ x +c_1&=-\frac {\left (\munderset {\textit {\_R} &=\operatorname {RootOf}\left (49 \textit {\_Z}^{7}-1\right )}{\sum }\frac {\ln \left (z -\textit {\_R} \right )}{\textit {\_R}^{6}}\right )}{343}+\frac {\left (\munderset {\textit {\_R} &=\operatorname {RootOf}\left (7 \textit {\_Z}^{7}+1\right )}{\sum }\frac {\ln \left (\sqrt {z}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{49}+\frac {\left (\munderset {\textit {\_R} &=\operatorname {RootOf}\left (7 \textit {\_Z}^{7}-1\right )}{\sum }\frac {\ln \left (\sqrt {z}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{49} \\ \end{align*}

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

Figure 79: Slope field plot
\(y^{\prime } = \left (\pi +x +7 y\right )^{{7}/{2}}\)

1.64.2 Solved using Lie symmetry for first order ode

Time used: 2.800 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=\left (\pi +x +7 y \right )^{{7}/{2}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}+\left (\pi +x +7 y \right )^{{7}/{2}} \left (b_{3}-a_{2}\right )-\left (\pi +x +7 y \right )^{7} a_{3}-\frac {7 \left (\pi +x +7 y \right )^{{5}/{2}} \left (x a_{2}+y a_{3}+a_{1}\right )}{2}-\frac {49 \left (\pi +x +7 y \right )^{{5}/{2}} \left (x b_{2}+y b_{3}+b_{1}\right )}{2} = 0 \end{equation}

Putting the above in normal form gives

\[ b_{2}-x^{7} a_{3}-823543 y^{7} a_{3}+\left (\pi +x +7 y \right )^{{7}/{2}} b_{3}-\left (\pi +x +7 y \right )^{{7}/{2}} a_{2}-\pi ^{7} a_{3}-\frac {7 \left (\pi +x +7 y \right )^{{5}/{2}} a_{1}}{2}-\frac {49 \left (\pi +x +7 y \right )^{{5}/{2}} b_{1}}{2}-49 x^{6} y a_{3}-1029 x^{5} y^{2} a_{3}-12005 x^{4} y^{3} a_{3}-84035 x^{3} y^{4} a_{3}-352947 x^{2} y^{5} a_{3}-823543 x \,y^{6} a_{3}-7 \pi ^{6} x a_{3}-49 \pi ^{6} y a_{3}-21 \pi ^{5} x^{2} a_{3}-1029 \pi ^{5} y^{2} a_{3}-35 \pi ^{4} x^{3} a_{3}-12005 \pi ^{4} y^{3} a_{3}-35 \pi ^{3} x^{4} a_{3}-84035 \pi ^{3} y^{4} a_{3}-21 \pi ^{2} x^{5} a_{3}-352947 \pi ^{2} y^{5} a_{3}-7 \pi \,x^{6} a_{3}-823543 \pi \,y^{6} a_{3}-\frac {7 \left (\pi +x +7 y \right )^{{5}/{2}} x a_{2}}{2}-\frac {7 \left (\pi +x +7 y \right )^{{5}/{2}} y a_{3}}{2}-\frac {49 \left (\pi +x +7 y \right )^{{5}/{2}} x b_{2}}{2}-\frac {49 \left (\pi +x +7 y \right )^{{5}/{2}} y b_{3}}{2}-980 \pi ^{3} x^{3} y a_{3}-10290 \pi ^{3} x^{2} y^{2} a_{3}-48020 \pi ^{3} x \,y^{3} a_{3}-735 \pi ^{2} x^{4} y a_{3}-10290 \pi ^{2} x^{3} y^{2} a_{3}-72030 \pi ^{2} x^{2} y^{3} a_{3}-252105 \pi ^{2} x \,y^{4} a_{3}-294 \pi \,x^{5} y a_{3}-5145 \pi \,x^{4} y^{2} a_{3}-48020 \pi \,x^{3} y^{3} a_{3}-252105 \pi \,x^{2} y^{4} a_{3}-705894 \pi x \,y^{5} a_{3}-294 \pi ^{5} x y a_{3}-735 \pi ^{4} x^{2} y a_{3}-5145 \pi ^{4} x \,y^{2} a_{3} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} 2 b_{2}-2 x^{7} a_{3}-1647086 y^{7} a_{3}+2 \left (\pi +x +7 y \right )^{{7}/{2}} b_{3}-2 \left (\pi +x +7 y \right )^{{7}/{2}} a_{2}-2 \pi ^{7} a_{3}-7 \left (\pi +x +7 y \right )^{{5}/{2}} a_{1}-49 \left (\pi +x +7 y \right )^{{5}/{2}} b_{1}-98 x^{6} y a_{3}-2058 x^{5} y^{2} a_{3}-24010 x^{4} y^{3} a_{3}-168070 x^{3} y^{4} a_{3}-705894 x^{2} y^{5} a_{3}-1647086 x \,y^{6} a_{3}-14 \pi ^{6} x a_{3}-98 \pi ^{6} y a_{3}-42 \pi ^{5} x^{2} a_{3}-2058 \pi ^{5} y^{2} a_{3}-70 \pi ^{4} x^{3} a_{3}-24010 \pi ^{4} y^{3} a_{3}-70 \pi ^{3} x^{4} a_{3}-168070 \pi ^{3} y^{4} a_{3}-42 \pi ^{2} x^{5} a_{3}-705894 \pi ^{2} y^{5} a_{3}-14 \pi \,x^{6} a_{3}-1647086 \pi \,y^{6} a_{3}-7 \left (\pi +x +7 y \right )^{{5}/{2}} x a_{2}-7 \left (\pi +x +7 y \right )^{{5}/{2}} y a_{3}-49 \left (\pi +x +7 y \right )^{{5}/{2}} x b_{2}-49 \left (\pi +x +7 y \right )^{{5}/{2}} y b_{3}-1960 \pi ^{3} x^{3} y a_{3}-20580 \pi ^{3} x^{2} y^{2} a_{3}-96040 \pi ^{3} x \,y^{3} a_{3}-1470 \pi ^{2} x^{4} y a_{3}-20580 \pi ^{2} x^{3} y^{2} a_{3}-144060 \pi ^{2} x^{2} y^{3} a_{3}-504210 \pi ^{2} x \,y^{4} a_{3}-588 \pi \,x^{5} y a_{3}-10290 \pi \,x^{4} y^{2} a_{3}-96040 \pi \,x^{3} y^{3} a_{3}-504210 \pi \,x^{2} y^{4} a_{3}-1411788 \pi x \,y^{5} a_{3}-588 \pi ^{5} x y a_{3}-1470 \pi ^{4} x^{2} y a_{3}-10290 \pi ^{4} x \,y^{2} a_{3} = 0 \end{equation}

Since the PDE has radicals, simplifying gives

\[ 2 b_{2}-2 x^{7} a_{3}-1647086 y^{7} a_{3}-2 \pi ^{7} a_{3}-182 \pi x \sqrt {\pi +x +7 y}\, y a_{2}-14 \pi x \sqrt {\pi +x +7 y}\, y a_{3}-686 \pi x \sqrt {\pi +x +7 y}\, y b_{2}-14 \pi x \sqrt {\pi +x +7 y}\, y b_{3}-9 x^{3} \sqrt {\pi +x +7 y}\, a_{2}-49 x^{3} \sqrt {\pi +x +7 y}\, b_{2}+2 x^{3} \sqrt {\pi +x +7 y}\, b_{3}-686 \sqrt {\pi +x +7 y}\, y^{3} a_{2}-343 \sqrt {\pi +x +7 y}\, y^{3} a_{3}-1715 \sqrt {\pi +x +7 y}\, y^{3} b_{3}-7 \pi ^{2} \sqrt {\pi +x +7 y}\, a_{1}-49 \pi ^{2} \sqrt {\pi +x +7 y}\, b_{1}-7 x^{2} \sqrt {\pi +x +7 y}\, a_{1}-49 x^{2} \sqrt {\pi +x +7 y}\, b_{1}-343 \sqrt {\pi +x +7 y}\, y^{2} a_{1}-2401 \sqrt {\pi +x +7 y}\, y^{2} b_{1}-2 \pi ^{3} \sqrt {\pi +x +7 y}\, a_{2}+2 \pi ^{3} \sqrt {\pi +x +7 y}\, b_{3}-98 x^{6} y a_{3}-2058 x^{5} y^{2} a_{3}-24010 x^{4} y^{3} a_{3}-168070 x^{3} y^{4} a_{3}-705894 x^{2} y^{5} a_{3}-1647086 x \,y^{6} a_{3}-14 \pi ^{6} x a_{3}-98 \pi ^{6} y a_{3}-42 \pi ^{5} x^{2} a_{3}-2058 \pi ^{5} y^{2} a_{3}-70 \pi ^{4} x^{3} a_{3}-24010 \pi ^{4} y^{3} a_{3}-70 \pi ^{3} x^{4} a_{3}-168070 \pi ^{3} y^{4} a_{3}-42 \pi ^{2} x^{5} a_{3}-705894 \pi ^{2} y^{5} a_{3}-14 \pi \,x^{6} a_{3}-1647086 \pi \,y^{6} a_{3}-7 x^{2} \sqrt {\pi +x +7 y}\, y a_{3}-686 x^{2} \sqrt {\pi +x +7 y}\, y b_{2}-7 x^{2} \sqrt {\pi +x +7 y}\, y b_{3}-637 x \sqrt {\pi +x +7 y}\, y^{2} a_{2}-98 x \sqrt {\pi +x +7 y}\, y^{2} a_{3}-2401 x \sqrt {\pi +x +7 y}\, y^{2} b_{2}-392 x \sqrt {\pi +x +7 y}\, y^{2} b_{3}-14 \pi x \sqrt {\pi +x +7 y}\, a_{1}-98 \pi x \sqrt {\pi +x +7 y}\, b_{1}-98 \pi \sqrt {\pi +x +7 y}\, y a_{1}-686 \pi \sqrt {\pi +x +7 y}\, y b_{1}-98 x \sqrt {\pi +x +7 y}\, y a_{1}-686 x \sqrt {\pi +x +7 y}\, y b_{1}-13 \pi ^{2} x \sqrt {\pi +x +7 y}\, a_{2}-49 \pi ^{2} x \sqrt {\pi +x +7 y}\, b_{2}+6 \pi ^{2} x \sqrt {\pi +x +7 y}\, b_{3}-42 \pi ^{2} \sqrt {\pi +x +7 y}\, y a_{2}-7 \pi ^{2} \sqrt {\pi +x +7 y}\, y a_{3}-7 \pi ^{2} \sqrt {\pi +x +7 y}\, y b_{3}-20 \pi \,x^{2} \sqrt {\pi +x +7 y}\, a_{2}-98 \pi \,x^{2} \sqrt {\pi +x +7 y}\, b_{2}+6 \pi \,x^{2} \sqrt {\pi +x +7 y}\, b_{3}-294 \pi \sqrt {\pi +x +7 y}\, y^{2} a_{2}-98 \pi \sqrt {\pi +x +7 y}\, y^{2} a_{3}-392 \pi \sqrt {\pi +x +7 y}\, y^{2} b_{3}-140 x^{2} \sqrt {\pi +x +7 y}\, y a_{2}-1960 \pi ^{3} x^{3} y a_{3}-20580 \pi ^{3} x^{2} y^{2} a_{3}-96040 \pi ^{3} x \,y^{3} a_{3}-1470 \pi ^{2} x^{4} y a_{3}-20580 \pi ^{2} x^{3} y^{2} a_{3}-144060 \pi ^{2} x^{2} y^{3} a_{3}-504210 \pi ^{2} x \,y^{4} a_{3}-588 \pi \,x^{5} y a_{3}-10290 \pi \,x^{4} y^{2} a_{3}-96040 \pi \,x^{3} y^{3} a_{3}-504210 \pi \,x^{2} y^{4} a_{3}-1411788 \pi x \,y^{5} a_{3}-588 \pi ^{5} x y a_{3}-1470 \pi ^{4} x^{2} y a_{3}-10290 \pi ^{4} x \,y^{2} a_{3} = 0 \]

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \left \{x, y, \sqrt {\pi +x +7 y}\right \} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \left \{x = v_{1}, y = v_{2}, \sqrt {\pi +x +7 y} = v_{3}\right \} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} -2 \pi ^{7} a_{3}-14 \pi ^{6} v_{1} a_{3}-98 \pi ^{6} v_{2} a_{3}-42 \pi ^{5} v_{1}^{2} a_{3}-588 \pi ^{5} v_{1} v_{2} a_{3}-2058 \pi ^{5} v_{2}^{2} a_{3}-70 \pi ^{4} v_{1}^{3} a_{3}-1470 \pi ^{4} v_{1}^{2} v_{2} a_{3}-10290 \pi ^{4} v_{1} v_{2}^{2} a_{3}-24010 \pi ^{4} v_{2}^{3} a_{3}-70 \pi ^{3} v_{1}^{4} a_{3}-1960 \pi ^{3} v_{1}^{3} v_{2} a_{3}-20580 \pi ^{3} v_{1}^{2} v_{2}^{2} a_{3}-96040 \pi ^{3} v_{1} v_{2}^{3} a_{3}-168070 \pi ^{3} v_{2}^{4} a_{3}-42 \pi ^{2} v_{1}^{5} a_{3}-1470 \pi ^{2} v_{1}^{4} v_{2} a_{3}-20580 \pi ^{2} v_{1}^{3} v_{2}^{2} a_{3}-144060 \pi ^{2} v_{1}^{2} v_{2}^{3} a_{3}-504210 \pi ^{2} v_{1} v_{2}^{4} a_{3}-705894 \pi ^{2} v_{2}^{5} a_{3}-14 \pi v_{1}^{6} a_{3}-588 \pi v_{1}^{5} v_{2} a_{3}-10290 \pi v_{1}^{4} v_{2}^{2} a_{3}-96040 \pi v_{1}^{3} v_{2}^{3} a_{3}-504210 \pi v_{1}^{2} v_{2}^{4} a_{3}-1411788 \pi v_{1} v_{2}^{5} a_{3}-1647086 \pi v_{2}^{6} a_{3}-2 v_{1}^{7} a_{3}-98 v_{1}^{6} v_{2} a_{3}-2058 v_{1}^{5} v_{2}^{2} a_{3}-24010 v_{1}^{4} v_{2}^{3} a_{3}-168070 v_{1}^{3} v_{2}^{4} a_{3}-705894 v_{1}^{2} v_{2}^{5} a_{3}-1647086 v_{1} v_{2}^{6} a_{3}-1647086 v_{2}^{7} a_{3}-2 \pi ^{3} v_{3} a_{2}+2 \pi ^{3} v_{3} b_{3}-13 \pi ^{2} v_{1} v_{3} a_{2}-42 \pi ^{2} v_{3} v_{2} a_{2}-7 \pi ^{2} v_{3} v_{2} a_{3}-49 \pi ^{2} v_{1} v_{3} b_{2}+6 \pi ^{2} v_{1} v_{3} b_{3}-7 \pi ^{2} v_{3} v_{2} b_{3}-20 \pi v_{1}^{2} v_{3} a_{2}-182 \pi v_{1} v_{3} v_{2} a_{2}-294 \pi v_{3} v_{2}^{2} a_{2}-14 \pi v_{1} v_{3} v_{2} a_{3}-98 \pi v_{3} v_{2}^{2} a_{3}-98 \pi v_{1}^{2} v_{3} b_{2}-686 \pi v_{1} v_{3} v_{2} b_{2}+6 \pi v_{1}^{2} v_{3} b_{3}-14 \pi v_{1} v_{3} v_{2} b_{3}-392 \pi v_{3} v_{2}^{2} b_{3}-9 v_{1}^{3} v_{3} a_{2}-140 v_{1}^{2} v_{3} v_{2} a_{2}-637 v_{1} v_{3} v_{2}^{2} a_{2}-686 v_{3} v_{2}^{3} a_{2}-7 v_{1}^{2} v_{3} v_{2} a_{3}-98 v_{1} v_{3} v_{2}^{2} a_{3}-343 v_{3} v_{2}^{3} a_{3}-49 v_{1}^{3} v_{3} b_{2}-686 v_{1}^{2} v_{3} v_{2} b_{2}-2401 v_{1} v_{3} v_{2}^{2} b_{2}+2 v_{1}^{3} v_{3} b_{3}-7 v_{1}^{2} v_{3} v_{2} b_{3}-392 v_{1} v_{3} v_{2}^{2} b_{3}-1715 v_{3} v_{2}^{3} b_{3}-7 \pi ^{2} v_{3} a_{1}-49 \pi ^{2} v_{3} b_{1}-14 \pi v_{1} v_{3} a_{1}-98 \pi v_{3} v_{2} a_{1}-98 \pi v_{1} v_{3} b_{1}-686 \pi v_{3} v_{2} b_{1}-7 v_{1}^{2} v_{3} a_{1}-98 v_{1} v_{3} v_{2} a_{1}-343 v_{3} v_{2}^{2} a_{1}-49 v_{1}^{2} v_{3} b_{1}-686 v_{1} v_{3} v_{2} b_{1}-2401 v_{3} v_{2}^{2} b_{1}+2 b_{2} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}, v_{3}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} 2 b_{2}-2 \pi ^{7} a_{3}+\left (-2 \pi ^{3} a_{2}+2 \pi ^{3} b_{3}-7 \pi ^{2} a_{1}-49 \pi ^{2} b_{1}\right ) v_{3}-2058 v_{1}^{5} v_{2}^{2} a_{3}-24010 v_{1}^{4} v_{2}^{3} a_{3}-168070 v_{1}^{3} v_{2}^{4} a_{3}-705894 v_{1}^{2} v_{2}^{5} a_{3}-1647086 v_{1} v_{2}^{6} a_{3}-14 \pi ^{6} v_{1} a_{3}-98 \pi ^{6} v_{2} a_{3}-42 \pi ^{5} v_{1}^{2} a_{3}-2058 \pi ^{5} v_{2}^{2} a_{3}-70 \pi ^{4} v_{1}^{3} a_{3}-24010 \pi ^{4} v_{2}^{3} a_{3}-70 \pi ^{3} v_{1}^{4} a_{3}-168070 \pi ^{3} v_{2}^{4} a_{3}-42 \pi ^{2} v_{1}^{5} a_{3}-705894 \pi ^{2} v_{2}^{5} a_{3}-14 \pi v_{1}^{6} a_{3}-1647086 \pi v_{2}^{6} a_{3}-98 v_{1}^{6} v_{2} a_{3}-2 v_{1}^{7} a_{3}-1647086 v_{2}^{7} a_{3}+\left (-13 \pi ^{2} a_{2}-49 \pi ^{2} b_{2}+6 \pi ^{2} b_{3}-14 \pi a_{1}-98 \pi b_{1}\right ) v_{1} v_{3}+\left (-686 a_{2}-343 a_{3}-1715 b_{3}\right ) v_{2}^{3} v_{3}+\left (-294 \pi a_{2}-98 \pi a_{3}-392 \pi b_{3}-343 a_{1}-2401 b_{1}\right ) v_{2}^{2} v_{3}+\left (-42 \pi ^{2} a_{2}-7 \pi ^{2} a_{3}-7 \pi ^{2} b_{3}-98 \pi a_{1}-686 \pi b_{1}\right ) v_{2} v_{3}+\left (-9 a_{2}-49 b_{2}+2 b_{3}\right ) v_{1}^{3} v_{3}+\left (-20 \pi a_{2}-98 \pi b_{2}+6 \pi b_{3}-7 a_{1}-49 b_{1}\right ) v_{1}^{2} v_{3}-1470 \pi ^{4} v_{1}^{2} v_{2} a_{3}-10290 \pi ^{4} v_{1} v_{2}^{2} a_{3}-1960 \pi ^{3} v_{1}^{3} v_{2} a_{3}-20580 \pi ^{3} v_{1}^{2} v_{2}^{2} a_{3}-96040 \pi ^{3} v_{1} v_{2}^{3} a_{3}-1470 \pi ^{2} v_{1}^{4} v_{2} a_{3}-20580 \pi ^{2} v_{1}^{3} v_{2}^{2} a_{3}-144060 \pi ^{2} v_{1}^{2} v_{2}^{3} a_{3}-504210 \pi ^{2} v_{1} v_{2}^{4} a_{3}-588 \pi v_{1}^{5} v_{2} a_{3}-10290 \pi v_{1}^{4} v_{2}^{2} a_{3}-96040 \pi v_{1}^{3} v_{2}^{3} a_{3}-504210 \pi v_{1}^{2} v_{2}^{4} a_{3}-1411788 \pi v_{1} v_{2}^{5} a_{3}-588 \pi ^{5} v_{1} v_{2} a_{3}+\left (-140 a_{2}-7 a_{3}-686 b_{2}-7 b_{3}\right ) v_{1}^{2} v_{2} v_{3}+\left (-637 a_{2}-98 a_{3}-2401 b_{2}-392 b_{3}\right ) v_{1} v_{2}^{2} v_{3}+\left (-182 \pi a_{2}-14 \pi a_{3}-686 \pi b_{2}-14 \pi b_{3}-98 a_{1}-686 b_{1}\right ) v_{1} v_{2} v_{3} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} -1647086 a_{3}&=0\\ -705894 a_{3}&=0\\ -168070 a_{3}&=0\\ -24010 a_{3}&=0\\ -2058 a_{3}&=0\\ -98 a_{3}&=0\\ -2 a_{3}&=0\\ -1647086 \pi a_{3}&=0\\ -1411788 \pi a_{3}&=0\\ -504210 \pi a_{3}&=0\\ -96040 \pi a_{3}&=0\\ -10290 \pi a_{3}&=0\\ -588 \pi a_{3}&=0\\ -14 \pi a_{3}&=0\\ -705894 \pi ^{2} a_{3}&=0\\ -504210 \pi ^{2} a_{3}&=0\\ -144060 \pi ^{2} a_{3}&=0\\ -20580 \pi ^{2} a_{3}&=0\\ -1470 \pi ^{2} a_{3}&=0\\ -42 \pi ^{2} a_{3}&=0\\ -168070 \pi ^{3} a_{3}&=0\\ -96040 \pi ^{3} a_{3}&=0\\ -20580 \pi ^{3} a_{3}&=0\\ -1960 \pi ^{3} a_{3}&=0\\ -70 \pi ^{3} a_{3}&=0\\ -24010 \pi ^{4} a_{3}&=0\\ -10290 \pi ^{4} a_{3}&=0\\ -1470 \pi ^{4} a_{3}&=0\\ -70 \pi ^{4} a_{3}&=0\\ -2058 \pi ^{5} a_{3}&=0\\ -588 \pi ^{5} a_{3}&=0\\ -42 \pi ^{5} a_{3}&=0\\ -98 \pi ^{6} a_{3}&=0\\ -14 \pi ^{6} a_{3}&=0\\ -686 a_{2}-343 a_{3}-1715 b_{3}&=0\\ -9 a_{2}-49 b_{2}+2 b_{3}&=0\\ -637 a_{2}-98 a_{3}-2401 b_{2}-392 b_{3}&=0\\ -140 a_{2}-7 a_{3}-686 b_{2}-7 b_{3}&=0\\ -2 \pi ^{7} a_{3}+2 b_{2}&=0\\ -2 \pi ^{3} a_{2}+2 \pi ^{3} b_{3}-7 \pi ^{2} a_{1}-49 \pi ^{2} b_{1}&=0\\ -294 \pi a_{2}-98 \pi a_{3}-392 \pi b_{3}-343 a_{1}-2401 b_{1}&=0\\ -42 \pi ^{2} a_{2}-7 \pi ^{2} a_{3}-7 \pi ^{2} b_{3}-98 \pi a_{1}-686 \pi b_{1}&=0\\ -20 \pi a_{2}-98 \pi b_{2}+6 \pi b_{3}-7 a_{1}-49 b_{1}&=0\\ -13 \pi ^{2} a_{2}-49 \pi ^{2} b_{2}+6 \pi ^{2} b_{3}-14 \pi a_{1}-98 \pi b_{1}&=0\\ -182 \pi a_{2}-14 \pi a_{3}-686 \pi b_{2}-14 \pi b_{3}-98 a_{1}-686 b_{1}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=-7 b_{1}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ b_{3}&=0 \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= -7 \\ \eta &= 1 \\ \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Therefore

\begin{align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {1}{-7}\\ &= -{\frac {1}{7}} \end{align*}

This is easily solved to give

\begin{align*} y = -\frac {x}{7}+c_1 \end{align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence

\begin{align*} R &= \frac {x}{7}+y \end{align*}

And \(S\) is found from

\begin{align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{-7} \end{align*}

Integrating gives

\begin{align*} S &= \int { \frac {dx}{T}}\\ &= -\frac {x}{7} \end{align*}

Where the constant of integration is set to zero as we just need one solution. Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= \left (\pi +x +7 y \right )^{{7}/{2}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= {\frac {1}{7}}\\ R_{y} &= 1\\ S_{x} &= -{\frac {1}{7}}\\ S_{y} &= 0 \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= -\frac {1}{1+7 \left (\pi +x +7 y \right )^{{7}/{2}}}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= -\frac {1}{1+7 \left (\pi +7 R \right )^{{7}/{2}}} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {1}{1+7 \left (\pi +7 R \right )^{{7}/{2}}}\, dR}\\ S \left (R \right ) &= -\frac {2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (7 \textit {\_Z}^{7}+1\right )}{\sum }\frac {\ln \left (\sqrt {\pi +7 R}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{343} + c_2 \end{align*}

\begin{align*} S \left (R \right )&= \int -\frac {1}{1+7 \left (\pi +7 R \right )^{{7}/{2}}}d R +c_2 \end{align*}

This results in

\begin{align*} -\frac {x}{7} = \int _{}^{y}-\frac {1}{1+7 \left (\pi +x +7 \textit {\_a} \right )^{{7}/{2}}}d \textit {\_a} +c_2 \end{align*}

Figure 80: Slope field plot
\(y^{\prime } = \left (\pi +x +7 y\right )^{{7}/{2}}\)

1.64.3 Solved as first order ode of type dAlembert

Time used: 13.093 (sec)

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} p = \left (\pi +x +7 y \right )^{{7}/{2}} \end{align*}

Solving for \(y\) from the above results in

\begin{align*} y &= \frac {p^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\tag {1A}\\ y &= \frac {\left (\cos \left (\frac {2 \pi }{7}\right )+i \cos \left (\frac {3 \pi }{14}\right )\right )^{2} p^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\tag {2A}\\ y &= \frac {\left (-\cos \left (\frac {3 \pi }{7}\right )+i \cos \left (\frac {\pi }{14}\right )\right )^{2} p^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\tag {3A}\\ y &= \frac {\left (-\cos \left (\frac {\pi }{7}\right )+i \cos \left (\frac {5 \pi }{14}\right )\right )^{2} p^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\tag {4A}\\ y &= \frac {\left (-\cos \left (\frac {\pi }{7}\right )-i \cos \left (\frac {5 \pi }{14}\right )\right )^{2} p^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\tag {5A}\\ y &= \frac {\left (-\cos \left (\frac {3 \pi }{7}\right )-i \cos \left (\frac {\pi }{14}\right )\right )^{2} p^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\tag {6A}\\ y &= \frac {\left (\cos \left (\frac {2 \pi }{7}\right )-i \cos \left (\frac {3 \pi }{14}\right )\right )^{2} p^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\tag {7A} \end{align*}

This has the form

\begin{align*} y=xf(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= -{\frac {1}{7}}\\ g &= \frac {p^{{2}/{7}}}{7}-\frac {\pi }{7} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {1}{7} = \frac {2 p^{\prime }\left (x \right )}{49 p^{{5}/{7}}}\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {1}{7} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=-{\frac {1}{7}} \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}}}{49}-\frac {\pi }{7}-\frac {x}{7} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {49 \left (p \left (x \right )+\frac {1}{7}\right ) p \left (x \right )^{{5}/{7}}}{2}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. Since initial conditions \(\left (x_0,p_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{p \left (x \right )}\frac {2}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau = x +c_1 \]

Singular solutions are found by solving

\begin{align*} \frac {7 \left (7 p +1\right ) p^{{5}/{7}}}{2}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -{\frac {1}{7}} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = \frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {2}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_1 \right )}^{{2}/{7}}}{7}-\frac {\pi }{7}-\frac {x}{7}\\ y = -\frac {\pi }{7}-\frac {x}{7}\\ y = \frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}}}{49}-\frac {\pi }{7}-\frac {x}{7}\\ \end{align*}

Solving ode 2A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= -{\frac {1}{7}}\\ g &= \frac {p^{{2}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )+i \sin \left (\frac {3 \pi }{7}\right )\right )}{7}-\frac {\pi }{7} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {1}{7} = \left (-\frac {2 \cos \left (\frac {3 \pi }{7}\right )}{49 p^{{5}/{7}}}+\frac {2 i \sin \left (\frac {3 \pi }{7}\right )}{49 p^{{5}/{7}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {1}{7} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=-{\frac {1}{7}} \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {7^{{5}/{7}} \left (-1\right )^{{6}/{7}}}{49}-\frac {\pi }{7}-\frac {x}{7} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {1}{7}}{-\frac {2 \cos \left (\frac {3 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}+\frac {2 i \sin \left (\frac {3 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. Since initial conditions \(\left (x_0,p_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{p \left (x \right )}\frac {2 \left (-1\right )^{{4}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} -\frac {7 \left (-1\right )^{{3}/{7}} \left (7 p +1\right ) p^{{5}/{7}}}{2}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -{\frac {1}{7}} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {2 \left (-1\right )^{{4}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_2 \right )}^{{2}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )+i \sin \left (\frac {3 \pi }{7}\right )\right )}{7}-\frac {\pi }{7}\\ y = -\frac {\pi }{7}-\frac {x}{7}\\ y = -\frac {x}{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )+i \sin \left (\frac {3 \pi }{7}\right )\right )}{49}-\frac {\pi }{7}\\ \end{align*}

Solving ode 3A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= -{\frac {1}{7}}\\ g &= \frac {p^{{2}/{7}} \left (-i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{7}-\frac {\pi }{7} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {1}{7} = \left (-\frac {2 i \sin \left (\frac {\pi }{7}\right )}{49 p^{{5}/{7}}}-\frac {2 \cos \left (\frac {\pi }{7}\right )}{49 p^{{5}/{7}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {1}{7} = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {1}{7}}{-\frac {2 i \sin \left (\frac {\pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}-\frac {2 \cos \left (\frac {\pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. Since initial conditions \(\left (x_0,p_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{p \left (x \right )}-\frac {2 \left (-1\right )^{{1}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} \frac {7 \left (-1\right )^{{6}/{7}} \left (7 p +1\right ) p^{{5}/{7}}}{2}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -{\frac {1}{7}} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {2 \left (-1\right )^{{1}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_3 \right )}^{{2}/{7}} \left (-i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{7}-\frac {\pi }{7}\\ y = -\frac {\pi }{7}-\frac {x}{7}\\ y = -\frac {x}{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (-i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{49}-\frac {\pi }{7}\\ \end{align*}

Solving ode 4A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= -{\frac {1}{7}}\\ g &= -\frac {\pi }{7}+\frac {p^{{2}/{7}} \left (-i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{7} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {1}{7} = \left (-\frac {2 i \sin \left (\frac {2 \pi }{7}\right )}{49 p^{{5}/{7}}}+\frac {2 \cos \left (\frac {2 \pi }{7}\right )}{49 p^{{5}/{7}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {1}{7} = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {1}{7}}{-\frac {2 i \sin \left (\frac {2 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}+\frac {2 \cos \left (\frac {2 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. Since initial conditions \(\left (x_0,p_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{p \left (x \right )}-\frac {2 \left (-1\right )^{{5}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} \frac {7 \left (-1\right )^{{2}/{7}} \left (7 p +1\right ) p^{{5}/{7}}}{2}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -{\frac {1}{7}} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = -\frac {x}{7}-\frac {\pi }{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {2 \left (-1\right )^{{5}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_4 \right )}^{{2}/{7}} \left (-i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{7}\\ y = -\frac {\pi }{7}-\frac {x}{7}\\ y = -\frac {x}{7}-\frac {\pi }{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (-i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{49}\\ \end{align*}

Solving ode 5A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= -{\frac {1}{7}}\\ g &= -\frac {\pi }{7}+\frac {p^{{2}/{7}} \left (i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{7} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {1}{7} = \left (\frac {2 i \sin \left (\frac {2 \pi }{7}\right )}{49 p^{{5}/{7}}}+\frac {2 \cos \left (\frac {2 \pi }{7}\right )}{49 p^{{5}/{7}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {1}{7} = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {1}{7}}{\frac {2 i \sin \left (\frac {2 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}+\frac {2 \cos \left (\frac {2 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. Since initial conditions \(\left (x_0,p_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{p \left (x \right )}\frac {2 \left (-1\right )^{{2}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau = x +c_5 \]

Singular solutions are found by solving

\begin{align*} -\frac {7 \left (-1\right )^{{5}/{7}} \left (7 p +1\right ) p^{{5}/{7}}}{2}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -{\frac {1}{7}} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = -\frac {x}{7}-\frac {\pi }{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {2 \left (-1\right )^{{2}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_5 \right )}^{{2}/{7}} \left (i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{7}\\ y = -\frac {\pi }{7}-\frac {x}{7}\\ y = -\frac {x}{7}-\frac {\pi }{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{49}\\ \end{align*}

Solving ode 6A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= -{\frac {1}{7}}\\ g &= \frac {p^{{2}/{7}} \left (i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{7}-\frac {\pi }{7} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {1}{7} = \left (\frac {2 i \sin \left (\frac {\pi }{7}\right )}{49 p^{{5}/{7}}}-\frac {2 \cos \left (\frac {\pi }{7}\right )}{49 p^{{5}/{7}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {1}{7} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=-{\frac {1}{7}} \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = -\frac {7^{{5}/{7}} \left (-1\right )^{{1}/{7}}}{49}-\frac {\pi }{7}-\frac {x}{7} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {1}{7}}{\frac {2 i \sin \left (\frac {\pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}-\frac {2 \cos \left (\frac {\pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. Since initial conditions \(\left (x_0,p_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{p \left (x \right )}\frac {2 \left (-1\right )^{{6}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau = x +c_6 \]

Singular solutions are found by solving

\begin{align*} -\frac {7 \left (-1\right )^{{1}/{7}} \left (7 p +1\right ) p^{{5}/{7}}}{2}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -{\frac {1}{7}} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {2 \left (-1\right )^{{6}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_6 \right )}^{{2}/{7}} \left (i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{7}-\frac {\pi }{7}\\ y = -\frac {\pi }{7}-\frac {x}{7}\\ y = -\frac {x}{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{49}-\frac {\pi }{7}\\ \end{align*}

Solving ode 7A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= -{\frac {1}{7}}\\ g &= \frac {p^{{2}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )-i \sin \left (\frac {3 \pi }{7}\right )\right )}{7}-\frac {\pi }{7} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {1}{7} = \left (-\frac {2 \cos \left (\frac {3 \pi }{7}\right )}{49 p^{{5}/{7}}}-\frac {2 i \sin \left (\frac {3 \pi }{7}\right )}{49 p^{{5}/{7}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {1}{7} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=-{\frac {1}{7}} \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = -\frac {\left (-7\right )^{{5}/{7}}}{49}-\frac {\pi }{7}-\frac {x}{7} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {1}{7}}{-\frac {2 \cos \left (\frac {3 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}-\frac {2 i \sin \left (\frac {3 \pi }{7}\right )}{49 p \left (x \right )^{{5}/{7}}}}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. Since initial conditions \(\left (x_0,p_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

\[ \int _{}^{p \left (x \right )}-\frac {2 \left (-1\right )^{{3}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau = x +c_7 \]

Singular solutions are found by solving

\begin{align*} \frac {7 \left (-1\right )^{{4}/{7}} \left (7 p +1\right ) p^{{5}/{7}}}{2}&= 0 \end{align*}

for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -{\frac {1}{7}} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {2 \left (-1\right )^{{3}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_7 \right )}^{{2}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )-i \sin \left (\frac {3 \pi }{7}\right )\right )}{7}-\frac {\pi }{7}\\ y = -\frac {\pi }{7}-\frac {x}{7}\\ y = -\frac {x}{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )-i \sin \left (\frac {3 \pi }{7}\right )\right )}{49}-\frac {\pi }{7}\\ \end{align*}

The solution

\[ y = -\frac {\pi }{7}-\frac {x}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {2 \left (-1\right )^{{4}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_2 \right )}^{{2}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )+i \sin \left (\frac {3 \pi }{7}\right )\right )}{7}-\frac {\pi }{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {\pi }{7}-\frac {x}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {2 \left (-1\right )^{{1}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_3 \right )}^{{2}/{7}} \left (-i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{7}-\frac {\pi }{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {\pi }{7}-\frac {x}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (-i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{49}-\frac {\pi }{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}-\frac {\pi }{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {2 \left (-1\right )^{{5}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_4 \right )}^{{2}/{7}} \left (-i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {\pi }{7}-\frac {x}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}-\frac {\pi }{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (-i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{49} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}-\frac {\pi }{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {2 \left (-1\right )^{{2}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_5 \right )}^{{2}/{7}} \left (i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {\pi }{7}-\frac {x}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}-\frac {\pi }{7}+\frac {\left (-1\right )^{{2}/{7}} 7^{{5}/{7}} \left (i \sin \left (\frac {2 \pi }{7}\right )+\cos \left (\frac {2 \pi }{7}\right )\right )}{49} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {2 \left (-1\right )^{{6}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_6 \right )}^{{2}/{7}} \left (i \sin \left (\frac {\pi }{7}\right )-\cos \left (\frac {\pi }{7}\right )\right )}{7}-\frac {\pi }{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {\pi }{7}-\frac {x}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {x}{7}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {2 \left (-1\right )^{{3}/{7}}}{7 \left (7 \tau +1\right ) \tau ^{{5}/{7}}}d \tau \right )+x +c_7 \right )}^{{2}/{7}} \left (-\cos \left (\frac {3 \pi }{7}\right )-i \sin \left (\frac {3 \pi }{7}\right )\right )}{7}-\frac {\pi }{7} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = -\frac {\pi }{7}-\frac {x}{7} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Figure 81: Slope field plot
\(y^{\prime } = \left (\pi +x +7 y\right )^{{7}/{2}}\)

1.64.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\pi +x +7 y\right )^{{7}/{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\pi +x +7 y\right )^{{7}/{2}} \end {array} \]

1.64.5 Maple trace

Methods for first order ODEs:
 

1.64.6 Maple dsolve solution

Solving time : 0.025 (sec)
Leaf size : 33

dsolve(diff(y(x),x) = (Pi+x+7*y(x))^(7/2), 
       y(x),singsol=all)
 

\[ y = -\frac {x}{7}+\operatorname {RootOf}\left (-x +7 \left (\int _{}^{\textit {\_Z}}\frac {1}{1+7 \left (\pi +7 \textit {\_a} \right )^{{7}/{2}}}d \textit {\_a} \right )+c_1 \right ) \]

1.64.7 Mathematica DSolve solution

Solving time : 30.453 (sec)
Leaf size : 43

DSolve[{D[y[x],x]==(Pi+x+7*y[x])^(7/2),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [-(7 y(x)+x+\pi ) \left (\operatorname {Hypergeometric2F1}\left (\frac {2}{7},1,\frac {9}{7},-7 (x+7 y(x)+\pi )^{7/2}\right )-1\right )-7 y(x)=c_1,y(x)\right ] \]