1.65 problem 65

Internal problem ID [8029]
Book : First order enumerated odes
Section : section 1
Problem number : 65
Date solved : Monday, October 21, 2024 at 04:44:14 PM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

Solve

\begin{align*} y^{\prime }&=\left (a +b x +c y\right )^{6} \end{align*}

1.65.1 Solved as first order homogeneous class C ode

Time used: 1.254 (sec)

Let

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

Then

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

Therefore

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

Hence the given ode can now be written as

\begin{align*} \frac {z^{\prime }\left (x \right )-b}{c}&=z^{6} \end{align*}

This is separable first order ode. Integrating

\begin{align*} \int d x&=\int \frac {1}{c \,z^{6}+b}d z \\ x +c_1&=\frac {\sqrt {3}\, \left (\frac {b}{c}\right )^{{1}/{6}} \ln \left (z^{2}+\sqrt {3}\, \left (\frac {b}{c}\right )^{{1}/{6}} z +\left (\frac {b}{c}\right )^{{1}/{3}}\right )}{12 b}+\frac {\left (\frac {b}{c}\right )^{{1}/{6}} \arctan \left (\frac {2 z}{\left (\frac {b}{c}\right )^{{1}/{6}}}+\sqrt {3}\right )}{6 b}+\frac {\left (\frac {b}{c}\right )^{{1}/{6}} \arctan \left (\frac {z}{\left (\frac {b}{c}\right )^{{1}/{6}}}\right )}{3 b}-\frac {\sqrt {3}\, \left (\frac {b}{c}\right )^{{1}/{6}} \ln \left (z^{2}-\sqrt {3}\, \left (\frac {b}{c}\right )^{{1}/{6}} z +\left (\frac {b}{c}\right )^{{1}/{3}}\right )}{12 b}+\frac {\left (\frac {b}{c}\right )^{{1}/{6}} \arctan \left (\frac {2 z}{\left (\frac {b}{c}\right )^{{1}/{6}}}-\sqrt {3}\right )}{6 b} \\ \end{align*}

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

1.65.2 Solved using Lie symmetry for first order ode

Time used: 1.496 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=\left (b x +c y +a \right )^{6}\\ 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 (b x +c y +a \right )^{6} \left (b_{3}-a_{2}\right )-\left (b x +c y +a \right )^{12} a_{3}-6 \left (b x +c y +a \right )^{5} b \left (x a_{2}+y a_{3}+a_{1}\right )-6 \left (b x +c y +a \right )^{5} c \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \text {Expression too large to display} \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} \text {Expression too large to display} \end{equation}

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

\[ \{x, y\} \]

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

\[ \{x = v_{1}, y = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} \text {Expression too large to display} \end{equation}

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

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

Equation (7E) now becomes

\begin{equation} \tag{8E} \text {Expression too large to display} \end{equation}

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

\begin{align*} -b^{12} a_{3}&=0\\ -c^{12} a_{3}&=0\\ -12 a \,b^{11} a_{3}&=0\\ -12 a \,c^{11} a_{3}&=0\\ -66 a^{2} b^{10} a_{3}&=0\\ -66 a^{2} c^{10} a_{3}&=0\\ -220 a^{3} b^{9} a_{3}&=0\\ -220 a^{3} c^{9} a_{3}&=0\\ -495 a^{4} b^{8} a_{3}&=0\\ -495 a^{4} c^{8} a_{3}&=0\\ -792 a^{5} b^{7} a_{3}&=0\\ -792 a^{5} c^{7} a_{3}&=0\\ -12 b \,c^{11} a_{3}&=0\\ -66 b^{2} c^{10} a_{3}&=0\\ -220 b^{3} c^{9} a_{3}&=0\\ -495 b^{4} c^{8} a_{3}&=0\\ -792 b^{5} c^{7} a_{3}&=0\\ -924 b^{6} c^{6} a_{3}&=0\\ -792 b^{7} c^{5} a_{3}&=0\\ -495 b^{8} c^{4} a_{3}&=0\\ -220 b^{9} c^{3} a_{3}&=0\\ -66 b^{10} c^{2} a_{3}&=0\\ -12 b^{11} c a_{3}&=0\\ -132 a b \,c^{10} a_{3}&=0\\ -660 a \,b^{2} c^{9} a_{3}&=0\\ -1980 a \,b^{3} c^{8} a_{3}&=0\\ -3960 a \,b^{4} c^{7} a_{3}&=0\\ -5544 a \,b^{5} c^{6} a_{3}&=0\\ -5544 a \,b^{6} c^{5} a_{3}&=0\\ -3960 a \,b^{7} c^{4} a_{3}&=0\\ -1980 a \,b^{8} c^{3} a_{3}&=0\\ -660 a \,b^{9} c^{2} a_{3}&=0\\ -132 a \,b^{10} c a_{3}&=0\\ -660 a^{2} b \,c^{9} a_{3}&=0\\ -2970 a^{2} b^{2} c^{8} a_{3}&=0\\ -7920 a^{2} b^{3} c^{7} a_{3}&=0\\ -13860 a^{2} b^{4} c^{6} a_{3}&=0\\ -16632 a^{2} b^{5} c^{5} a_{3}&=0\\ -13860 a^{2} b^{6} c^{4} a_{3}&=0\\ -7920 a^{2} b^{7} c^{3} a_{3}&=0\\ -2970 a^{2} b^{8} c^{2} a_{3}&=0\\ -660 a^{2} b^{9} c a_{3}&=0\\ -1980 a^{3} b \,c^{8} a_{3}&=0\\ -7920 a^{3} b^{2} c^{7} a_{3}&=0\\ -18480 a^{3} b^{3} c^{6} a_{3}&=0\\ -27720 a^{3} b^{4} c^{5} a_{3}&=0\\ -27720 a^{3} b^{5} c^{4} a_{3}&=0\\ -18480 a^{3} b^{6} c^{3} a_{3}&=0\\ -7920 a^{3} b^{7} c^{2} a_{3}&=0\\ -1980 a^{3} b^{8} c a_{3}&=0\\ -3960 a^{4} b \,c^{7} a_{3}&=0\\ -13860 a^{4} b^{2} c^{6} a_{3}&=0\\ -27720 a^{4} b^{3} c^{5} a_{3}&=0\\ -34650 a^{4} b^{4} c^{4} a_{3}&=0\\ -27720 a^{4} b^{5} c^{3} a_{3}&=0\\ -13860 a^{4} b^{6} c^{2} a_{3}&=0\\ -3960 a^{4} b^{7} c a_{3}&=0\\ -5544 a^{5} b \,c^{6} a_{3}&=0\\ -16632 a^{5} b^{2} c^{5} a_{3}&=0\\ -27720 a^{5} b^{3} c^{4} a_{3}&=0\\ -27720 a^{5} b^{4} c^{3} a_{3}&=0\\ -16632 a^{5} b^{5} c^{2} a_{3}&=0\\ -5544 a^{5} b^{6} c a_{3}&=0\\ -5544 a^{6} b^{5} c a_{3}-6 b^{6} a_{3}-36 b^{5} c a_{2}-30 b^{4} c^{2} b_{2}&=0\\ -924 a^{6} c^{6} a_{3}-6 b \,c^{5} a_{3}-c^{6} a_{2}-5 c^{6} b_{3}&=0\\ -924 a^{6} b^{6} a_{3}-7 b^{6} a_{2}+b^{6} b_{3}-6 b^{5} c b_{2}&=0\\ -12 a^{11} c a_{3}-6 a^{5} b a_{3}-6 a^{5} c a_{2}-30 a^{4} b c a_{1}-30 a^{4} c^{2} b_{1}&=0\\ -5544 a^{6} b \,c^{5} a_{3}-30 b^{2} c^{4} a_{3}-12 b \,c^{5} a_{2}-24 b \,c^{5} b_{3}-6 c^{6} b_{2}&=0\\ -13860 a^{6} b^{2} c^{4} a_{3}-60 b^{3} c^{3} a_{3}-45 b^{2} c^{4} a_{2}-45 b^{2} c^{4} b_{3}-30 b \,c^{5} b_{2}&=0\\ -18480 a^{6} b^{3} c^{3} a_{3}-60 b^{4} c^{2} a_{3}-80 b^{3} c^{3} a_{2}-40 b^{3} c^{3} b_{3}-60 b^{2} c^{4} b_{2}&=0\\ -13860 a^{6} b^{4} c^{2} a_{3}-30 b^{5} c a_{3}-75 b^{4} c^{2} a_{2}-15 b^{4} c^{2} b_{3}-60 b^{3} c^{3} b_{2}&=0\\ -3960 a^{7} b^{4} c a_{3}-30 a \,b^{5} a_{3}-150 a \,b^{4} c a_{2}-120 a \,b^{3} c^{2} b_{2}-30 b^{5} c a_{1}-30 b^{4} c^{2} b_{1}&=0\\ -1980 a^{8} b^{3} c a_{3}-60 a^{2} b^{4} a_{3}-240 a^{2} b^{3} c a_{2}-180 a^{2} b^{2} c^{2} b_{2}-120 a \,b^{4} c a_{1}-120 a \,b^{3} c^{2} b_{1}&=0\\ -660 a^{9} b^{2} c a_{3}-60 a^{3} b^{3} a_{3}-180 a^{3} b^{2} c a_{2}-120 a^{3} b \,c^{2} b_{2}-180 a^{2} b^{3} c a_{1}-180 a^{2} b^{2} c^{2} b_{1}&=0\\ -132 a^{10} b c a_{3}-30 a^{4} b^{2} a_{3}-60 a^{4} b c a_{2}-30 a^{4} c^{2} b_{2}-120 a^{3} b^{2} c a_{1}-120 a^{3} b \,c^{2} b_{1}&=0\\ -792 a^{7} c^{5} a_{3}-30 a b \,c^{4} a_{3}-6 a \,c^{5} a_{2}-24 a \,c^{5} b_{3}-6 b \,c^{5} a_{1}-6 c^{6} b_{1}&=0\\ -495 a^{8} c^{4} a_{3}-60 a^{2} b \,c^{3} a_{3}-15 a^{2} c^{4} a_{2}-45 a^{2} c^{4} b_{3}-30 a b \,c^{4} a_{1}-30 a \,c^{5} b_{1}&=0\\ -220 a^{9} c^{3} a_{3}-60 a^{3} b \,c^{2} a_{3}-20 a^{3} c^{3} a_{2}-40 a^{3} c^{3} b_{3}-60 a^{2} b \,c^{3} a_{1}-60 a^{2} c^{4} b_{1}&=0\\ -66 a^{10} c^{2} a_{3}-30 a^{4} b c a_{3}-15 a^{4} c^{2} a_{2}-15 a^{4} c^{2} b_{3}-60 a^{3} b \,c^{2} a_{1}-60 a^{3} c^{3} b_{1}&=0\\ -792 a^{7} b^{5} a_{3}-36 a \,b^{5} a_{2}+6 a \,b^{5} b_{3}-30 a \,b^{4} c b_{2}-6 b^{6} a_{1}-6 b^{5} c b_{1}&=0\\ -495 a^{8} b^{4} a_{3}-75 a^{2} b^{4} a_{2}+15 a^{2} b^{4} b_{3}-60 a^{2} b^{3} c b_{2}-30 a \,b^{5} a_{1}-30 a \,b^{4} c b_{1}&=0\\ -220 a^{9} b^{3} a_{3}-80 a^{3} b^{3} a_{2}+20 a^{3} b^{3} b_{3}-60 a^{3} b^{2} c b_{2}-60 a^{2} b^{4} a_{1}-60 a^{2} b^{3} c b_{1}&=0\\ -66 a^{10} b^{2} a_{3}-45 a^{4} b^{2} a_{2}+15 a^{4} b^{2} b_{3}-30 a^{4} b c b_{2}-60 a^{3} b^{3} a_{1}-60 a^{3} b^{2} c b_{1}&=0\\ -12 a^{11} b a_{3}-12 a^{5} b a_{2}+6 a^{5} b b_{3}-6 a^{5} c b_{2}-30 a^{4} b^{2} a_{1}-30 a^{4} b c b_{1}&=0\\ -a^{12} a_{3}-a^{6} a_{2}+a^{6} b_{3}-6 a^{5} b a_{1}-6 a^{5} c b_{1}+b_{2}&=0\\ -3960 a^{7} b \,c^{4} a_{3}-120 a \,b^{2} c^{3} a_{3}-60 a b \,c^{4} a_{2}-90 a b \,c^{4} b_{3}-30 a \,c^{5} b_{2}-30 b^{2} c^{4} a_{1}-30 b \,c^{5} b_{1}&=0\\ -7920 a^{7} b^{2} c^{3} a_{3}-180 a \,b^{3} c^{2} a_{3}-180 a \,b^{2} c^{3} a_{2}-120 a \,b^{2} c^{3} b_{3}-120 a b \,c^{4} b_{2}-60 b^{3} c^{3} a_{1}-60 b^{2} c^{4} b_{1}&=0\\ -7920 a^{7} b^{3} c^{2} a_{3}-120 a \,b^{4} c a_{3}-240 a \,b^{3} c^{2} a_{2}-60 a \,b^{3} c^{2} b_{3}-180 a \,b^{2} c^{3} b_{2}-60 b^{4} c^{2} a_{1}-60 b^{3} c^{3} b_{1}&=0\\ -1980 a^{8} b \,c^{3} a_{3}-180 a^{2} b^{2} c^{2} a_{3}-120 a^{2} b \,c^{3} a_{2}-120 a^{2} b \,c^{3} b_{3}-60 a^{2} c^{4} b_{2}-120 a \,b^{2} c^{3} a_{1}-120 a b \,c^{4} b_{1}&=0\\ -2970 a^{8} b^{2} c^{2} a_{3}-180 a^{2} b^{3} c a_{3}-270 a^{2} b^{2} c^{2} a_{2}-90 a^{2} b^{2} c^{2} b_{3}-180 a^{2} b \,c^{3} b_{2}-180 a \,b^{3} c^{2} a_{1}-180 a \,b^{2} c^{3} b_{1}&=0\\ -660 a^{9} b \,c^{2} a_{3}-120 a^{3} b^{2} c a_{3}-120 a^{3} b \,c^{2} a_{2}-60 a^{3} b \,c^{2} b_{3}-60 a^{3} c^{3} b_{2}-180 a^{2} b^{2} c^{2} a_{1}-180 a^{2} b \,c^{3} b_{1}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=-\frac {c b_{1}}{b}\\ 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 &= -\frac {c}{b} \\ \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}{-\frac {c}{b}}\\ &= -\frac {b}{c} \end{align*}

This is easily solved to give

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

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

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

And \(S\) is found from

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

Integrating gives

\begin{align*} S &= \int { \frac {dx}{T}}\\ &= -\frac {b x}{c} \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 (b x +c y +a \right )^{6} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= \frac {b}{c}\\ R_{y} &= 1\\ S_{x} &= -\frac {b}{c}\\ 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 {b}{c \left (\frac {b}{c}+\left (b x +c y +a \right )^{6}\right )}\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 {b}{R^{6} c^{7}+6 R^{5} a \,c^{6}+15 R^{4} a^{2} c^{5}+20 R^{3} a^{3} c^{4}+15 R^{2} a^{4} c^{3}+6 R \,a^{5} c^{2}+a^{6} c +b} \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 {b}{R^{6} c^{7}+6 R^{5} a \,c^{6}+15 R^{4} a^{2} c^{5}+20 R^{3} a^{3} c^{4}+15 R^{2} a^{4} c^{3}+6 R \,a^{5} c^{2}+a^{6} c +b}\, dR}\\ S \left (R \right ) &= -\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{7} \textit {\_Z}^{6}+6 \textit {\_Z}^{5} a \,c^{6}+15 \textit {\_Z}^{4} a^{2} c^{5}+20 \textit {\_Z}^{3} a^{3} c^{4}+15 \textit {\_Z}^{2} a^{4} c^{3}+6 a^{5} c^{2} \textit {\_Z} +a^{6} c +b \right )}{\sum }\frac {\ln \left (R -\textit {\_R} \right )}{\textit {\_R}^{5} c^{5}+5 \textit {\_R}^{4} a \,c^{4}+10 \textit {\_R}^{3} a^{2} c^{3}+10 \textit {\_R}^{2} a^{3} c^{2}+5 \textit {\_R} \,a^{4} c +a^{5}}\right )}{6 c^{2}} + c_2 \end{align*}

\begin{align*} S \left (R \right )&= \int -\frac {b}{R^{6} c^{7}+6 R^{5} a \,c^{6}+15 R^{4} a^{2} c^{5}+20 R^{3} a^{3} c^{4}+15 R^{2} a^{4} c^{3}+6 R \,a^{5} c^{2}+a^{6} c +b}d R +c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} -\frac {b x}{c} = \int _{}^{\frac {c y+b x}{c}}-\frac {b}{\textit {\_a}^{6} c^{7}+6 \textit {\_a}^{5} a \,c^{6}+15 \textit {\_a}^{4} a^{2} c^{5}+20 \textit {\_a}^{3} a^{3} c^{4}+15 \textit {\_a}^{2} a^{4} c^{3}+6 \textit {\_a} \,a^{5} c^{2}+a^{6} c +b}d \textit {\_a} +c_2 \end{align*}

1.65.3 Solved as first order ode of type dAlembert

Time used: 2.426 (sec)

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

\begin{align*} p = \left (b x +c y +a \right )^{6} \end{align*}

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

\begin{align*} y &= -\frac {b x}{c}+\frac {p^{{1}/{6}}-a}{c}\tag {1A}\\ y &= -\frac {b x}{c}+\frac {\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) p^{{1}/{6}}-a}{c}\tag {2A}\\ y &= -\frac {b x}{c}+\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) p^{{1}/{6}}-a}{c}\tag {3A}\\ y &= -\frac {b x}{c}+\frac {-p^{{1}/{6}}-a}{c}\tag {4A}\\ y &= -\frac {b x}{c}+\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) p^{{1}/{6}}-a}{c}\tag {5A}\\ y &= -\frac {b x}{c}+\frac {\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) p^{{1}/{6}}-a}{c}\tag {6A} \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 {b}{c}\\ g &= \frac {p^{{1}/{6}}-a}{c} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {b}{c} = \frac {p^{\prime }\left (x \right )}{6 p^{{5}/{6}} c}\tag {2A} \end{align*}

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

\begin{align*} p +\frac {b}{c} = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} y = \frac {-b x +\left (-\frac {b}{c}\right )^{{1}/{6}}-a}{c} \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 ) = 6 \left (p \left (x \right )+\frac {b}{c}\right ) p \left (x \right )^{{5}/{6}} c\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

\[ \int _{}^{p \left (x \right )}\frac {1}{6 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau = x +c_1 \]

Singular solutions are found by solving

\begin{align*} 6 \left (p c +b \right ) p^{{5}/{6}}&= 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 {b}{c} \end{align*}

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

\begin{align*} y = -\frac {b x}{c}+\frac {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{6 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_1 \right )}^{{1}/{6}}-a}{c}\\ y = -\frac {b x}{c}-\frac {a}{c}\\ y = -\frac {b x}{c}+\frac {\left (-\frac {b}{c}\right )^{{1}/{6}}-a}{c}\\ \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 {b}{c}\\ g &= \frac {i p^{{1}/{6}} \sqrt {3}+p^{{1}/{6}}-2 a}{2 c} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {b}{c} = \left (\frac {i \sqrt {3}}{12 c \,p^{{5}/{6}}}+\frac {1}{12 p^{{5}/{6}} c}\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 {b}{c} = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} y = \frac {i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}-2 b x +\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c} \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 {b}{c}}{\frac {i \sqrt {3}}{12 c p \left (x \right )^{{5}/{6}}}+\frac {1}{12 p \left (x \right )^{{5}/{6}} c}}\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

\[ \int _{}^{p \left (x \right )}\frac {1+i \sqrt {3}}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \frac {12 \left (p c +b \right ) p^{{5}/{6}}}{1+i \sqrt {3}}&= 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 {b}{c} \end{align*}

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

\begin{align*} y = -\frac {b x}{c}+\frac {i {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1+i \sqrt {3}}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_2 \right )}^{{1}/{6}} \sqrt {3}+{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1+i \sqrt {3}}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_2 \right )}^{{1}/{6}}-2 a}{2 c}\\ y = -\frac {b x}{c}-\frac {a}{c}\\ y = -\frac {b x}{c}+\frac {i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}+\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c}\\ \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 {b}{c}\\ g &= \frac {i p^{{1}/{6}} \sqrt {3}-p^{{1}/{6}}-2 a}{2 c} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {b}{c} = \left (\frac {i \sqrt {3}}{12 c \,p^{{5}/{6}}}-\frac {1}{12 p^{{5}/{6}} c}\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 {b}{c} = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} y = \frac {i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}-2 b x -\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c} \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 {b}{c}}{\frac {i \sqrt {3}}{12 c p \left (x \right )^{{5}/{6}}}-\frac {1}{12 p \left (x \right )^{{5}/{6}} c}}\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

\[ \int _{}^{p \left (x \right )}\frac {i \sqrt {3}-1}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} \frac {12 \left (p c +b \right ) p^{{5}/{6}}}{i \sqrt {3}-1}&= 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 {b}{c} \end{align*}

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

\begin{align*} y = -\frac {b x}{c}+\frac {i {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {i \sqrt {3}-1}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_3 \right )}^{{1}/{6}} \sqrt {3}-{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {i \sqrt {3}-1}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_3 \right )}^{{1}/{6}}-2 a}{2 c}\\ y = -\frac {b x}{c}-\frac {a}{c}\\ y = -\frac {b x}{c}+\frac {i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}-\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c}\\ \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 {b}{c}\\ g &= \frac {-p^{{1}/{6}}-a}{c} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {b}{c} = -\frac {p^{\prime }\left (x \right )}{6 p^{{5}/{6}} c}\tag {2A} \end{align*}

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

\begin{align*} p +\frac {b}{c} = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} y = \frac {-b x -\left (-\frac {b}{c}\right )^{{1}/{6}}-a}{c} \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 ) = -6 \left (p \left (x \right )+\frac {b}{c}\right ) p \left (x \right )^{{5}/{6}} c\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

\[ \int _{}^{p \left (x \right )}-\frac {1}{6 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} -6 \left (p c +b \right ) p^{{5}/{6}}&= 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 {b}{c} \end{align*}

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

\begin{align*} y = -\frac {b x}{c}+\frac {-{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {1}{6 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_4 \right )}^{{1}/{6}}-a}{c}\\ y = -\frac {b x}{c}-\frac {a}{c}\\ y = -\frac {b x}{c}+\frac {-\left (-\frac {b}{c}\right )^{{1}/{6}}-a}{c}\\ \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 {b}{c}\\ g &= \frac {-i p^{{1}/{6}} \sqrt {3}-p^{{1}/{6}}-2 a}{2 c} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {b}{c} = \left (-\frac {i \sqrt {3}}{12 c \,p^{{5}/{6}}}-\frac {1}{12 p^{{5}/{6}} c}\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 {b}{c} = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} y = \frac {-i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}-2 b x -\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c} \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 {b}{c}}{-\frac {i \sqrt {3}}{12 c p \left (x \right )^{{5}/{6}}}-\frac {1}{12 p \left (x \right )^{{5}/{6}} c}}\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

\[ \int _{}^{p \left (x \right )}-\frac {1+i \sqrt {3}}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau = x +c_5 \]

Singular solutions are found by solving

\begin{align*} -\frac {12 \left (p c +b \right ) p^{{5}/{6}}}{1+i \sqrt {3}}&= 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 {b}{c} \end{align*}

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

\begin{align*} y = -\frac {b x}{c}+\frac {-i {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {1+i \sqrt {3}}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_5 \right )}^{{1}/{6}} \sqrt {3}-{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {1+i \sqrt {3}}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_5 \right )}^{{1}/{6}}-2 a}{2 c}\\ y = -\frac {b x}{c}-\frac {a}{c}\\ y = -\frac {b x}{c}+\frac {-i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}-\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c}\\ \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 {b}{c}\\ g &= \frac {-i p^{{1}/{6}} \sqrt {3}+p^{{1}/{6}}-2 a}{2 c} \end{align*}

Hence (2) becomes

\begin{align*} p +\frac {b}{c} = \left (-\frac {i \sqrt {3}}{12 c \,p^{{5}/{6}}}+\frac {1}{12 p^{{5}/{6}} c}\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 {b}{c} = 0 \end{align*}

Solving the above for \(p\) results in

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

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

\begin{align*} y = \frac {-i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}-2 b x +\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c} \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 {b}{c}}{-\frac {i \sqrt {3}}{12 c p \left (x \right )^{{5}/{6}}}+\frac {1}{12 p \left (x \right )^{{5}/{6}} c}}\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

\[ \int _{}^{p \left (x \right )}-\frac {i \sqrt {3}-1}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau = x +c_6 \]

Singular solutions are found by solving

\begin{align*} -\frac {12 \left (p c +b \right ) p^{{5}/{6}}}{i \sqrt {3}-1}&= 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 {b}{c} \end{align*}

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

\begin{align*} y = -\frac {b x}{c}+\frac {-i {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {i \sqrt {3}-1}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_6 \right )}^{{1}/{6}} \sqrt {3}+{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {i \sqrt {3}-1}{12 \left (c \tau +b \right ) \tau ^{{5}/{6}}}d \tau \right )+x +c_6 \right )}^{{1}/{6}}-2 a}{2 c}\\ y = -\frac {b x}{c}-\frac {a}{c}\\ y = -\frac {b x}{c}+\frac {-i \left (-\frac {b}{c}\right )^{{1}/{6}} \sqrt {3}+\left (-\frac {b}{c}\right )^{{1}/{6}}-2 a}{2 c}\\ \end{align*}

The solution

\[ y = -\frac {b x}{c}-\frac {a}{c} \]

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

\[ y = -\frac {b x}{c}-\frac {a}{c} \]

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

\[ y = -\frac {b x}{c}-\frac {a}{c} \]

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

\[ y = -\frac {b x}{c}-\frac {a}{c} \]

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

\[ y = -\frac {b x}{c}-\frac {a}{c} \]

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

\[ y = -\frac {b x}{c}-\frac {a}{c} \]

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

1.65.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (a +b x +c y\right )^{6} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (a +b x +c y\right )^{6} \end {array} \]

1.65.5 Maple trace

Methods for first order ODEs:
 

1.65.6 Maple dsolve solution

Solving time : 0.016 (sec)
Leaf size : 94

dsolve(diff(y(x),x) = (a+b*x+c*y(x))^6, 
       y(x),singsol=all)
 

\[ y = \frac {\operatorname {RootOf}\left (\left (\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a}^{6} c^{7}+6 \textit {\_a}^{5} a \,c^{6}+15 \textit {\_a}^{4} a^{2} c^{5}+20 \textit {\_a}^{3} a^{3} c^{4}+15 \textit {\_a}^{2} a^{4} c^{3}+6 \textit {\_a} \,a^{5} c^{2}+a^{6} c +b}d \textit {\_a} \right ) c -x +c_1 \right ) c -b x}{c} \]

1.65.7 Mathematica DSolve solution

Solving time : 1.783 (sec)
Leaf size : 274

DSolve[{D[y[x],x]==(a+b*x+c*y[x])^6,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\frac {-4 \sqrt [6]{b} \arctan \left (\frac {\sqrt [6]{c} (a+b x+c y(x))}{\sqrt [6]{b}}\right )+2 \sqrt [6]{b} \arctan \left (\sqrt {3}-\frac {2 \sqrt [6]{c} (a+b x+c y(x))}{\sqrt [6]{b}}\right )-2 \sqrt [6]{b} \arctan \left (\frac {2 \sqrt [6]{c} (a+b x+c y(x))}{\sqrt [6]{b}}+\sqrt {3}\right )+\sqrt {3} \sqrt [6]{b} \log \left (\sqrt [3]{c} (a+b x+c y(x))^2-\sqrt {3} \sqrt [6]{b} \sqrt [6]{c} (a+b x+c y(x))+\sqrt [3]{b}\right )-\sqrt {3} \sqrt [6]{b} \log \left (\sqrt [3]{c} (a+b x+c y(x))^2+\sqrt {3} \sqrt [6]{b} \sqrt [6]{c} (a+b x+c y(x))+\sqrt [3]{b}\right )+12 a \sqrt [6]{c}+12 b \sqrt [6]{c} x+12 c^{7/6} y(x)}{12 b \sqrt [6]{c}}-\frac {c y(x)}{b}=c_1,y(x)\right ] \]