2.1.9 Problem 9

Internal problem ID [4084]
Book : Differential equations, Shepley L. Ross, 1964
Section : 2.4, page 55
Problem number : 9
Date solved : Saturday, December 06, 2025 at 04:16:56 PM
CAS classification : [[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]

2.1.9.1 Existence and uniqueness analysis

\begin{align*} 3 x -y \left (x \right )-6+\left (x +y \left (x \right )+2\right ) y^{\prime }\left (x \right )&=0 \\ y \left (2\right ) &= -2 \\ \end{align*}

This is non linear first order ODE. In canonical form it is written as

\begin{align*} y^{\prime }\left (x \right ) &= f(x,y \left (x \right ))\\ &= \frac {-3 x +y +6}{x +y +2} \end{align*}

The \(x\) domain of \(f(x,y \left (x \right ))\) when \(y=-2\) is

\[ \{x <0\boldsymbol {\lor }0<x\} \]

And the point \(x_0 = 2\) is inside this domain. The \(y\) domain of \(f(x,y \left (x \right ))\) when \(x=2\) is

\[ \{y <-4\boldsymbol {\lor }-4<y\} \]

And the point \(y_0 = -2\) is inside this domain. Now we will look at the continuity of

\begin{align*} \frac {\partial f}{\partial y} &= \frac {\partial }{\partial y}\left (\frac {-3 x +y +6}{x +y +2}\right ) \\ &= \frac {1}{x +y +2}-\frac {-3 x +y +6}{\left (x +y +2\right )^{2}} \end{align*}

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=-2\) is

\[ \{x <0\boldsymbol {\lor }0<x\} \]

And the point \(x_0 = 2\) is inside this domain. The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=2\) is

\[ \{y <-4\boldsymbol {\lor }-4<y\} \]

And the point \(y_0 = -2\) is inside this domain. Therefore solution exists and is unique.

2.1.9.2 Solved using first_order_ode_homog_type_maple_C

0.475 (sec)

Entering first order ode homog type maple C solver

\begin{align*} 3 x -y-6+\left (x +y+2\right ) y^{\prime }&=0 \\ y \left (2\right ) &= -2 \\ \end{align*}

Let \(Y = y -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)

\[ \frac {d}{d X}Y \left (X \right ) = \frac {-3 X -3 x_{0} +Y \left (X \right )+y_{0} +6}{X +x_{0} +Y \left (X \right )+y_{0} +2} \]

Solving for possible values of \(x_{0}\) and \(y_{0}\) which makes the above ode a homogeneous ode results in

\begin{align*} x_{0}&=1\\ y_{0}&=-3 \end{align*}

Using these values now it is possible to easily solve for \(Y \left (X \right )\). The above ode now becomes

\begin{align*} \frac {d}{d X}Y \left (X \right ) = \frac {-3 X +Y \left (X \right )}{X +Y \left (X \right )} \end{align*}

In canonical form, the ODE is

\begin{align*} Y' &= F(X,Y)\\ &= \frac {-3 X +Y}{X +Y}\tag {1} \end{align*}

An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if

\[ f(t^n X, t^n Y)= t^n f(X,Y) \]

In this case, it can be seen that both \(M=-3 X +Y\) and \(N=X +Y\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence

\[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \]

Applying the transformation \(Y=uX\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u &= \frac {u -3}{u +1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {u \left (X \right )-3}{u \left (X \right )+1}-u \left (X \right )}{X} \end{align*}

Or

\[ \frac {d}{d X}u \left (X \right )-\frac {\frac {u \left (X \right )-3}{u \left (X \right )+1}-u \left (X \right )}{X} = 0 \]

Or

\[ \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )+u \left (X \right )^{2}+\left (\frac {d}{d X}u \left (X \right )\right ) X +3 = 0 \]

Or

\[ X \left (u \left (X \right )+1\right ) \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}+3 = 0 \]

Which is now solved as separable in \(u \left (X \right )\).

The ode

\begin{equation} \frac {d}{d X}u \left (X \right ) = -\frac {u \left (X \right )^{2}+3}{X \left (u \left (X \right )+1\right )} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d X}u \left (X \right )&= -\frac {u \left (X \right )^{2}+3}{X \left (u \left (X \right )+1\right )}\\ &= f(X) g(u) \end{align*}

Where

\begin{align*} f(X) &= -\frac {1}{X}\\ g(u) &= \frac {u^{2}+3}{u +1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\ \int { \frac {u +1}{u^{2}+3}\,du} &= \int { -\frac {1}{X} \,dX} \\ \end{align*}

\[ \frac {\ln \left (u \left (X \right )^{2}+3\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {u \left (X \right ) \sqrt {3}}{3}\right )}{3}=\ln \left (\frac {1}{X}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \frac {u^{2}+3}{u +1}=0 \]

for \(u \left (X \right )\) gives

\begin{align*} u \left (X \right )&=-i \sqrt {3}\\ u \left (X \right )&=i \sqrt {3} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (u \left (X \right )^{2}+3\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {u \left (X \right ) \sqrt {3}}{3}\right )}{3} &= \ln \left (\frac {1}{X}\right )+c_1 \\ u \left (X \right ) &= -i \sqrt {3} \\ u \left (X \right ) &= i \sqrt {3} \\ \end{align*}

Converting \(\frac {\ln \left (u \left (X \right )^{2}+3\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {u \left (X \right ) \sqrt {3}}{3}\right )}{3} = \ln \left (\frac {1}{X}\right )+c_1\) back to \(Y \left (X \right )\) gives

\begin{align*} \frac {\sqrt {3}\, \left (\sqrt {3}\, \ln \left (\frac {Y \left (X \right )^{2}+3 X^{2}}{X^{2}}\right )+2 \arctan \left (\frac {Y \left (X \right ) \sqrt {3}}{3 X}\right )\right )}{6} = \ln \left (\frac {1}{X}\right )+c_1 \end{align*}

Converting \(u \left (X \right ) = -i \sqrt {3}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = -i \sqrt {3}\, X \end{align*}

Converting \(u \left (X \right ) = i \sqrt {3}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = i \sqrt {3}\, X \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} \frac {\sqrt {3}\, \left (\sqrt {3}\, \ln \left (\frac {Y \left (X \right )^{2}+3 X^{2}}{X^{2}}\right )+2 \arctan \left (\frac {Y \left (X \right ) \sqrt {3}}{3 X}\right )\right )}{6} = \ln \left (\frac {1}{X}\right )+c_1\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}

Or

\begin{align*} Y &= y -3\\ X &= 1+x \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} \frac {\sqrt {3}\, \left (\sqrt {3}\, \ln \left (\frac {\left (y+3\right )^{2}+3 \left (-1+x \right )^{2}}{\left (-1+x \right )^{2}}\right )+2 \arctan \left (\frac {\left (y+3\right ) \sqrt {3}}{-3+3 x}\right )\right )}{6} = \ln \left (\frac {1}{-1+x}\right )+c_1 \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = -i \sqrt {3}\, X\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}

Or

\begin{align*} Y &= y -3\\ X &= 1+x \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} y+3 = -i \sqrt {3}\, \left (-1+x \right ) \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = i \sqrt {3}\, X\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}

Or

\begin{align*} Y &= y -3\\ X &= 1+x \end{align*}

Then the solution in \(y\) becomes using EQ (A)

\begin{align*} y+3 = i \sqrt {3}\, \left (-1+x \right ) \end{align*}

Simplifying the above gives

\begin{align*} \frac {\sqrt {3}\, \left (\sqrt {3}\, \ln \left (\frac {\left (y+3\right )^{2}+3 \left (-1+x \right )^{2}}{\left (-1+x \right )^{2}}\right )+2 \arctan \left (\frac {\left (y+3\right ) \sqrt {3}}{-3+3 x}\right )\right )}{6} &= \ln \left (\frac {1}{-1+x}\right )+c_1 \\ y+3 &= -i \sqrt {3}\, \left (-1+x \right ) \\ y+3 &= i \sqrt {3}\, \left (-1+x \right ) \\ \end{align*}

Solving for initial conditions the solution is

\begin{align*} \frac {\sqrt {3}\, \left (\sqrt {3}\, \ln \left (\frac {\left (y+3\right )^{2}+3 \left (-1+x \right )^{2}}{\left (-1+x \right )^{2}}\right )+2 \arctan \left (\frac {\left (y+3\right ) \sqrt {3}}{-3+3 x}\right )\right )}{6} &= \ln \left (\frac {1}{-1+x}\right )+\frac {\sqrt {3}\, \pi }{18}+\ln \left (2\right ) \\ y+3 &= -i \sqrt {3}\, \left (-1+x \right ) \\ y+3 &= i \sqrt {3}\, \left (-1+x \right ) \\ \end{align*}

The solution

\[ y+3 = -i \sqrt {3}\, \left (-1+x \right ) \]

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

\[ y+3 = i \sqrt {3}\, \left (-1+x \right ) \]

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

Figure 2.26: Slope field \(3 x -y-6+\left (x +y+2\right ) y^{\prime } = 0\)

Summary of solutions found

\begin{align*} \frac {\sqrt {3}\, \left (\sqrt {3}\, \ln \left (\frac {\left (y+3\right )^{2}+3 \left (-1+x \right )^{2}}{\left (-1+x \right )^{2}}\right )+2 \arctan \left (\frac {\left (y+3\right ) \sqrt {3}}{-3+3 x}\right )\right )}{6} &= \ln \left (\frac {1}{-1+x}\right )+\frac {\sqrt {3}\, \pi }{18}+\ln \left (2\right ) \\ \end{align*}

Entering first order ode abel second kind solver

\begin{align*} 3 x -y-6+\left (x +y+2\right ) y^{\prime }&=0 \\ y \left (2\right ) &= -2 \\ \end{align*}

Applying transformation

\begin{align*} y&=\frac {1}{u(x)}-g \end{align*}

Results in the new ode which is Abel first kind

\begin{align*} 3 x -x u \left (x \right )-6+\left (x +x u \left (x \right )+2\right ) \left (u \left (x \right )+x u^{\prime }\left (x \right )\right ) = 0 \end{align*}

Which is now solved Unknown ode type.

2.1.9.3 Solved using first_order_ode_abel_second_kind_solved_by_converting_to_first_kind

7.655 (sec) This is Abel second kind ODE, it has the form

\[ \left (y \left (x \right )+g\right )y^{\prime }\left (x \right )= f_0(x)+f_1(x) y \left (x \right ) +f_2(x)y \left (x \right )^{2}+f_3(x)y \left (x \right )^{3} \]

Comparing the above to given ODE which is

\begin{align*}3 x -y \left (x \right )-6+\left (x +y \left (x \right )+2\right ) y^{\prime }\left (x \right ) = 0\tag {1} \end{align*}

Shows that

\begin{align*} g &= 2+x\\ f_0 &= -3 x +6\\ f_1 &= 1\\ f_2 &= 0\\ f_3 &= 0 \end{align*}

Applying transformation

\begin{align*} y \left (x \right )&=\frac {1}{u(x)}-g \end{align*}

Results in the new ode which is Abel first kind

\begin{align*} u^{\prime }\left (x \right ) = \left (4 x -4\right ) u \left (x \right )^{3}-2 u \left (x \right )^{2} \end{align*}

Which is now solved.

Entering first order ode abel first kind solverThis is Abel first kind ODE, it has the form

\[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \]

Comparing the above to given ODE which is

\begin{align*}u^{\prime }\left (x \right )&=\left (4 x -4\right ) u \left (x \right )^{3}-2 u \left (x \right )^{2}\tag {1} \end{align*}

Therefore

\begin{align*} f_0 &= 0\\ f_1 &= 0\\ f_2 &= -2\\ f_3 &= 4 x -4 \end{align*}

Hence

\begin{align*} f'_{0} &= 0\\ f'_{3} &= 4 \end{align*}

Since \(f_2(x)=-2\) is not zero, then the followingtransformation is used to remove \(f_2\). Let \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\) or

\begin{align*} u \left (x \right ) &= u(x) - \left ( \frac {-2}{12 x -12} \right ) \\ &= u \left (x \right )+\frac {1}{-6+6 x} \end{align*}

The above transformation applied to (1) gives a new ODE as

\begin{align*} u^{\prime }\left (x \right ) = \frac {\left (216 x^{3}-648 x^{2}+648 x -216\right ) u \left (x \right )^{3}}{54 \left (-1+x \right )^{2}}+\frac {\left (-18 x +18\right ) u \left (x \right )}{54 \left (-1+x \right )^{2}}+\frac {7}{54 \left (-1+x \right )^{2}}\tag {2} \end{align*}

The above ODE (2) can now be solved.

Entering first order ode LIE solverWriting the ode as

\begin{align*} u^{\prime }\left (x \right )&=\frac {216 u^{3} x^{3}-648 u^{3} x^{2}+648 u^{3} x -216 u^{3}-18 u x +18 u +7}{54 \left (-1+x \right )^{2}}\\ u^{\prime }\left (x \right )&= \omega \left ( x,u \left (x \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{u \left (x \right )}-\xi _{x}\right ) -\omega ^{2}\xi _{u \left (x \right )}-\omega _{x}\xi -\omega _{u \left (x \right )}\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 &= u a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= u b_{3}+x b_{2}+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}+\frac {\left (216 u^{3} x^{3}-648 u^{3} x^{2}+648 u^{3} x -216 u^{3}-18 u x +18 u +7\right ) \left (b_{3}-a_{2}\right )}{54 \left (-1+x \right )^{2}}-\frac {\left (216 u^{3} x^{3}-648 u^{3} x^{2}+648 u^{3} x -216 u^{3}-18 u x +18 u +7\right )^{2} a_{3}}{2916 \left (-1+x \right )^{4}}-\left (\frac {648 u^{3} x^{2}-1296 u^{3} x +648 u^{3}-18 u}{54 \left (-1+x \right )^{2}}-\frac {216 u^{3} x^{3}-648 u^{3} x^{2}+648 u^{3} x -216 u^{3}-18 u x +18 u +7}{27 \left (-1+x \right )^{3}}\right ) \left (u a_{3}+x a_{2}+a_{1}\right )-\frac {\left (648 u^{2} x^{3}-1944 u^{2} x^{2}+1944 u^{2} x -648 u^{2}-18 x +18\right ) \left (u b_{3}+x b_{2}+b_{1}\right )}{54 \left (-1+x \right )^{2}} = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {46656 u^{6} x^{6} a_{3}-279936 u^{6} x^{5} a_{3}+699840 u^{6} x^{4} a_{3}-933120 u^{6} x^{3} a_{3}+699840 u^{6} x^{2} a_{3}+3888 u^{4} x^{4} a_{3}+23328 u^{3} x^{5} a_{2}+23328 u^{3} x^{5} b_{3}+34992 u^{2} x^{6} b_{2}-279936 u^{6} x a_{3}-15552 u^{4} x^{3} a_{3}+11664 u^{3} x^{4} a_{1}-104976 u^{3} x^{4} a_{2}-116640 u^{3} x^{4} b_{3}+34992 u^{2} x^{5} b_{1}-174960 u^{2} x^{5} b_{2}+46656 u^{6} a_{3}+23328 u^{4} x^{2} a_{3}-46656 u^{3} x^{3} a_{1}+186624 u^{3} x^{3} a_{2}+3024 u^{3} x^{3} a_{3}+233280 u^{3} x^{3} b_{3}-174960 u^{2} x^{4} b_{1}+349920 u^{2} x^{4} b_{2}-15552 u^{4} x a_{3}+69984 u^{3} x^{2} a_{1}-163296 u^{3} x^{2} a_{2}-9072 u^{3} x^{2} a_{3}-233280 u^{3} x^{2} b_{3}+349920 u^{2} x^{3} b_{1}-349920 u^{2} x^{3} b_{2}+3888 u^{4} a_{3}-46656 u^{3} x a_{1}+69984 u^{3} x a_{2}+9072 u^{3} x a_{3}+116640 u^{3} x b_{3}+1296 u^{2} x^{2} a_{3}-349920 u^{2} x^{2} b_{1}+174960 u^{2} x^{2} b_{2}-3888 x^{4} b_{2}+11664 u^{3} a_{1}-11664 u^{3} a_{2}-3024 u^{3} a_{3}-23328 u^{3} b_{3}-2592 u^{2} x a_{3}+174960 u^{2} x b_{1}-34992 u^{2} x b_{2}+972 u \,x^{2} a_{1}+972 u \,x^{2} a_{2}-972 x^{3} b_{1}+14580 x^{3} b_{2}+1296 u^{2} a_{3}-34992 u^{2} b_{1}-1944 u x a_{1}-1944 u x a_{2}-1008 u x a_{3}-378 x^{2} a_{2}+2916 x^{2} b_{1}-20412 x^{2} b_{2}-378 x^{2} b_{3}+972 u a_{1}+972 u a_{2}+1008 u a_{3}-756 x a_{1}-2916 x b_{1}+12636 x b_{2}+756 x b_{3}+756 a_{1}+378 a_{2}+49 a_{3}+972 b_{1}-2916 b_{2}-378 b_{3}}{2916 \left (-1+x \right )^{4}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -46656 u^{6} x^{6} a_{3}+279936 u^{6} x^{5} a_{3}-699840 u^{6} x^{4} a_{3}+933120 u^{6} x^{3} a_{3}-699840 u^{6} x^{2} a_{3}-3888 u^{4} x^{4} a_{3}-23328 u^{3} x^{5} a_{2}-23328 u^{3} x^{5} b_{3}-34992 u^{2} x^{6} b_{2}+279936 u^{6} x a_{3}+15552 u^{4} x^{3} a_{3}-11664 u^{3} x^{4} a_{1}+104976 u^{3} x^{4} a_{2}+116640 u^{3} x^{4} b_{3}-34992 u^{2} x^{5} b_{1}+174960 u^{2} x^{5} b_{2}-46656 u^{6} a_{3}-23328 u^{4} x^{2} a_{3}+46656 u^{3} x^{3} a_{1}-186624 u^{3} x^{3} a_{2}-3024 u^{3} x^{3} a_{3}-233280 u^{3} x^{3} b_{3}+174960 u^{2} x^{4} b_{1}-349920 u^{2} x^{4} b_{2}+15552 u^{4} x a_{3}-69984 u^{3} x^{2} a_{1}+163296 u^{3} x^{2} a_{2}+9072 u^{3} x^{2} a_{3}+233280 u^{3} x^{2} b_{3}-349920 u^{2} x^{3} b_{1}+349920 u^{2} x^{3} b_{2}-3888 u^{4} a_{3}+46656 u^{3} x a_{1}-69984 u^{3} x a_{2}-9072 u^{3} x a_{3}-116640 u^{3} x b_{3}-1296 u^{2} x^{2} a_{3}+349920 u^{2} x^{2} b_{1}-174960 u^{2} x^{2} b_{2}+3888 x^{4} b_{2}-11664 u^{3} a_{1}+11664 u^{3} a_{2}+3024 u^{3} a_{3}+23328 u^{3} b_{3}+2592 u^{2} x a_{3}-174960 u^{2} x b_{1}+34992 u^{2} x b_{2}-972 u \,x^{2} a_{1}-972 u \,x^{2} a_{2}+972 x^{3} b_{1}-14580 x^{3} b_{2}-1296 u^{2} a_{3}+34992 u^{2} b_{1}+1944 u x a_{1}+1944 u x a_{2}+1008 u x a_{3}+378 x^{2} a_{2}-2916 x^{2} b_{1}+20412 x^{2} b_{2}+378 x^{2} b_{3}-972 u a_{1}-972 u a_{2}-1008 u a_{3}+756 x a_{1}+2916 x b_{1}-12636 x b_{2}-756 x b_{3}-756 a_{1}-378 a_{2}-49 a_{3}-972 b_{1}+2916 b_{2}+378 b_{3} = 0 \end{equation}

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

\[ \{u, x\} \]

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

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

The above PDE (6E) now becomes

\begin{equation} \tag{7E} -46656 a_{3} v_{1}^{6} v_{2}^{6}+279936 a_{3} v_{1}^{6} v_{2}^{5}-699840 a_{3} v_{1}^{6} v_{2}^{4}+933120 a_{3} v_{1}^{6} v_{2}^{3}-23328 a_{2} v_{1}^{3} v_{2}^{5}-699840 a_{3} v_{1}^{6} v_{2}^{2}-3888 a_{3} v_{1}^{4} v_{2}^{4}-34992 b_{2} v_{1}^{2} v_{2}^{6}-23328 b_{3} v_{1}^{3} v_{2}^{5}-11664 a_{1} v_{1}^{3} v_{2}^{4}+104976 a_{2} v_{1}^{3} v_{2}^{4}+279936 a_{3} v_{1}^{6} v_{2}+15552 a_{3} v_{1}^{4} v_{2}^{3}-34992 b_{1} v_{1}^{2} v_{2}^{5}+174960 b_{2} v_{1}^{2} v_{2}^{5}+116640 b_{3} v_{1}^{3} v_{2}^{4}+46656 a_{1} v_{1}^{3} v_{2}^{3}-186624 a_{2} v_{1}^{3} v_{2}^{3}-46656 a_{3} v_{1}^{6}-23328 a_{3} v_{1}^{4} v_{2}^{2}-3024 a_{3} v_{1}^{3} v_{2}^{3}+174960 b_{1} v_{1}^{2} v_{2}^{4}-349920 b_{2} v_{1}^{2} v_{2}^{4}-233280 b_{3} v_{1}^{3} v_{2}^{3}-69984 a_{1} v_{1}^{3} v_{2}^{2}+163296 a_{2} v_{1}^{3} v_{2}^{2}+15552 a_{3} v_{1}^{4} v_{2}+9072 a_{3} v_{1}^{3} v_{2}^{2}-349920 b_{1} v_{1}^{2} v_{2}^{3}+349920 b_{2} v_{1}^{2} v_{2}^{3}+233280 b_{3} v_{1}^{3} v_{2}^{2}+46656 a_{1} v_{1}^{3} v_{2}-69984 a_{2} v_{1}^{3} v_{2}-3888 a_{3} v_{1}^{4}-9072 a_{3} v_{1}^{3} v_{2}-1296 a_{3} v_{1}^{2} v_{2}^{2}+349920 b_{1} v_{1}^{2} v_{2}^{2}-174960 b_{2} v_{1}^{2} v_{2}^{2}+3888 b_{2} v_{2}^{4}-116640 b_{3} v_{1}^{3} v_{2}-11664 a_{1} v_{1}^{3}-972 a_{1} v_{1} v_{2}^{2}+11664 a_{2} v_{1}^{3}-972 a_{2} v_{1} v_{2}^{2}+3024 a_{3} v_{1}^{3}+2592 a_{3} v_{1}^{2} v_{2}-174960 b_{1} v_{1}^{2} v_{2}+972 b_{1} v_{2}^{3}+34992 b_{2} v_{1}^{2} v_{2}-14580 b_{2} v_{2}^{3}+23328 b_{3} v_{1}^{3}+1944 a_{1} v_{1} v_{2}+1944 a_{2} v_{1} v_{2}+378 a_{2} v_{2}^{2}-1296 a_{3} v_{1}^{2}+1008 a_{3} v_{1} v_{2}+34992 b_{1} v_{1}^{2}-2916 b_{1} v_{2}^{2}+20412 b_{2} v_{2}^{2}+378 b_{3} v_{2}^{2}-972 a_{1} v_{1}+756 a_{1} v_{2}-972 a_{2} v_{1}-1008 a_{3} v_{1}+2916 b_{1} v_{2}-12636 b_{2} v_{2}-756 b_{3} v_{2}-756 a_{1}-378 a_{2}-49 a_{3}-972 b_{1}+2916 b_{2}+378 b_{3} = 0 \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} 15552 a_{3} v_{1}^{4} v_{2}+\left (-11664 a_{1}+11664 a_{2}+3024 a_{3}+23328 b_{3}\right ) v_{1}^{3}+\left (-1296 a_{3}+34992 b_{1}\right ) v_{1}^{2}+\left (-972 a_{1}-972 a_{2}-1008 a_{3}\right ) v_{1}+\left (972 b_{1}-14580 b_{2}\right ) v_{2}^{3}+\left (378 a_{2}-2916 b_{1}+20412 b_{2}+378 b_{3}\right ) v_{2}^{2}+\left (756 a_{1}+2916 b_{1}-12636 b_{2}-756 b_{3}\right ) v_{2}-46656 a_{3} v_{1}^{6} v_{2}^{6}+279936 a_{3} v_{1}^{6} v_{2}^{5}-699840 a_{3} v_{1}^{6} v_{2}^{4}+933120 a_{3} v_{1}^{6} v_{2}^{3}-699840 a_{3} v_{1}^{6} v_{2}^{2}-3888 a_{3} v_{1}^{4} v_{2}^{4}-34992 b_{2} v_{1}^{2} v_{2}^{6}+279936 a_{3} v_{1}^{6} v_{2}+15552 a_{3} v_{1}^{4} v_{2}^{3}-23328 a_{3} v_{1}^{4} v_{2}^{2}-756 a_{1}-378 a_{2}-49 a_{3}-972 b_{1}+2916 b_{2}+378 b_{3}+\left (-23328 a_{2}-23328 b_{3}\right ) v_{1}^{3} v_{2}^{5}+\left (-11664 a_{1}+104976 a_{2}+116640 b_{3}\right ) v_{1}^{3} v_{2}^{4}+\left (46656 a_{1}-186624 a_{2}-3024 a_{3}-233280 b_{3}\right ) v_{1}^{3} v_{2}^{3}+\left (-69984 a_{1}+163296 a_{2}+9072 a_{3}+233280 b_{3}\right ) v_{1}^{3} v_{2}^{2}+\left (46656 a_{1}-69984 a_{2}-9072 a_{3}-116640 b_{3}\right ) v_{1}^{3} v_{2}+\left (-34992 b_{1}+174960 b_{2}\right ) v_{1}^{2} v_{2}^{5}+\left (174960 b_{1}-349920 b_{2}\right ) v_{1}^{2} v_{2}^{4}+\left (-349920 b_{1}+349920 b_{2}\right ) v_{1}^{2} v_{2}^{3}+\left (-1296 a_{3}+349920 b_{1}-174960 b_{2}\right ) v_{1}^{2} v_{2}^{2}+\left (2592 a_{3}-174960 b_{1}+34992 b_{2}\right ) v_{1}^{2} v_{2}+\left (-972 a_{1}-972 a_{2}\right ) v_{1} v_{2}^{2}+\left (1944 a_{1}+1944 a_{2}+1008 a_{3}\right ) v_{1} v_{2}-46656 a_{3} v_{1}^{6}-3888 a_{3} v_{1}^{4}+3888 b_{2} v_{2}^{4} = 0 \end{equation}

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

\begin{align*} -699840 a_{3}&=0\\ -46656 a_{3}&=0\\ -23328 a_{3}&=0\\ -3888 a_{3}&=0\\ 15552 a_{3}&=0\\ 279936 a_{3}&=0\\ 933120 a_{3}&=0\\ -34992 b_{2}&=0\\ 3888 b_{2}&=0\\ -972 a_{1}-972 a_{2}&=0\\ -23328 a_{2}-23328 b_{3}&=0\\ -1296 a_{3}+34992 b_{1}&=0\\ -349920 b_{1}+349920 b_{2}&=0\\ -34992 b_{1}+174960 b_{2}&=0\\ 972 b_{1}-14580 b_{2}&=0\\ 174960 b_{1}-349920 b_{2}&=0\\ -11664 a_{1}+104976 a_{2}+116640 b_{3}&=0\\ -972 a_{1}-972 a_{2}-1008 a_{3}&=0\\ 1944 a_{1}+1944 a_{2}+1008 a_{3}&=0\\ -1296 a_{3}+349920 b_{1}-174960 b_{2}&=0\\ 2592 a_{3}-174960 b_{1}+34992 b_{2}&=0\\ -69984 a_{1}+163296 a_{2}+9072 a_{3}+233280 b_{3}&=0\\ -11664 a_{1}+11664 a_{2}+3024 a_{3}+23328 b_{3}&=0\\ 756 a_{1}+2916 b_{1}-12636 b_{2}-756 b_{3}&=0\\ 46656 a_{1}-186624 a_{2}-3024 a_{3}-233280 b_{3}&=0\\ 46656 a_{1}-69984 a_{2}-9072 a_{3}-116640 b_{3}&=0\\ 378 a_{2}-2916 b_{1}+20412 b_{2}+378 b_{3}&=0\\ -756 a_{1}-378 a_{2}-49 a_{3}-972 b_{1}+2916 b_{2}+378 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

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

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,u\right ) \xi \\ &= u - \left (\frac {216 u^{3} x^{3}-648 u^{3} x^{2}+648 u^{3} x -216 u^{3}-18 u x +18 u +7}{54 \left (-1+x \right )^{2}}\right ) \left (1-x\right ) \\ &= \frac {7+216 \left (-1+x \right )^{3} u^{3}+\left (-36+36 x \right ) u}{-54+54 x}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,u\right ) \to \left ( R,S \right )\) 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 u}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial u}\right ) S(x,u) = 1\). 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 \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {7+216 \left (-1+x \right )^{3} u^{3}+\left (-36+36 x \right ) u}{-54+54 x}}} dy \end{align*}

Which results in

\begin{align*} S&= \left (-54+54 x \right ) \left (\frac {\left (6-6 x \right ) \ln \left (36 u^{2} x^{2}-72 u^{2} x +36 u^{2}-6 u x +6 u +7\right )}{648 x^{2}-1296 x +648}+\frac {\left (2-\frac {\left (6-6 x \right )^{2}}{2 \left (36 x^{2}-72 x +36\right )}\right ) \sqrt {3}\, \arctan \left (\frac {\left (2 \left (36 x^{2}-72 x +36\right ) u +6-6 x \right ) \sqrt {3}}{-54+54 x}\right )}{-243+243 x}+\frac {\ln \left (6 u x -6 u +1\right )}{-54+54 x}\right ) \end{align*}

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,u) S_{u} }{ R_{x} + \omega (x,u) R_{u} }\tag {2} \end{align*}

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

\begin{align*} \omega (x,u) &= \frac {216 u^{3} x^{3}-648 u^{3} x^{2}+648 u^{3} x -216 u^{3}-18 u x +18 u +7}{54 \left (-1+x \right )^{2}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{u} &= 0\\ S_{x} &= \frac {u}{4 \left (\frac {7}{36}+\left (-1+x \right )^{2} u^{2}+\frac {\left (1-x \right ) u}{6}\right ) \left (\frac {1}{6}+\left (-1+x \right ) u \right )}\\ S_{u} &= \frac {-1+x}{4 \left (\frac {7}{36}+\left (-1+x \right )^{2} u^{2}+\frac {\left (1-x \right ) u}{6}\right ) \left (\frac {1}{6}+\left (-1+x \right ) u \right )} \end{align*}

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

\begin{align*} \frac {dS}{dR} &= \frac {1}{-1+x}\tag {2A} \end{align*}

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

\begin{align*} \frac {dS}{dR} &= \frac {1}{-1+R} \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+R}\, dR}\\ S \left (R \right ) &= \ln \left (-1+R \right ) + c_2 \end{align*}

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

\begin{align*} \frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+12 \left (-1+x \right ) u \left (x \right )\right ) \sqrt {3}}{9}\right )}{3}-\frac {\ln \left (7+36 \left (-1+x \right )^{2} u \left (x \right )^{2}+\left (6-6 x \right ) u \left (x \right )\right )}{2}+\ln \left (u \left (x \right ) \left (-6+6 x \right )+1\right ) = \ln \left (-1+x \right )+c_2 \end{align*}

Simplifying the above gives

\begin{align*} \frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+12 \left (-1+x \right ) u \left (x \right )\right ) \sqrt {3}}{9}\right )}{3}-\frac {\ln \left (7+36 \left (-1+x \right )^{2} u \left (x \right )^{2}+\left (6-6 x \right ) u \left (x \right )\right )}{2}+\ln \left (1+6 \left (-1+x \right ) u \left (x \right )\right ) &= \ln \left (-1+x \right )+c_2 \\ \end{align*}

Substituting \(u=u \left (x \right )-\frac {2}{3 \left (4 x -4\right )}\) in the above solution gives

\begin{align*} \frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+12 \left (-1+x \right ) \left (u \left (x \right )-\frac {2}{3 \left (4 x -4\right )}\right )\right ) \sqrt {3}}{9}\right )}{3}-\frac {\ln \left (7+36 \left (-1+x \right )^{2} \left (u \left (x \right )-\frac {2}{3 \left (4 x -4\right )}\right )^{2}+\left (6-6 x \right ) \left (u \left (x \right )-\frac {2}{3 \left (4 x -4\right )}\right )\right )}{2}+\ln \left (1+6 \left (-1+x \right ) \left (u \left (x \right )-\frac {2}{3 \left (4 x -4\right )}\right )\right ) = \ln \left (-1+x \right )+c_2 \end{align*}

Simplifying the above gives

\begin{align*} \frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+\left (4 x -4\right ) u \left (x \right )\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (1+4 \left (-1+x \right )^{2} u \left (x \right )^{2}+\left (-2 x +2\right ) u \left (x \right )\right )}{2}+\ln \left (2\right )+\ln \left (u \left (x \right ) \left (-1+x \right )\right ) &= \ln \left (-1+x \right )+c_2 \\ \end{align*}

Substituting \(u \left (x \right )=\frac {1}{x +y \left (x \right )+2}\) in the above solution gives

\[ \frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+\frac {4 x -4}{x +y \left (x \right )+2}\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (1+\frac {4 \left (-1+x \right )^{2}}{\left (x +y \left (x \right )+2\right )^{2}}+\frac {-2 x +2}{x +y \left (x \right )+2}\right )}{2}+\ln \left (2\right )+\ln \left (\frac {-1+x}{x +y \left (x \right )+2}\right ) = \ln \left (-1+x \right )+c_2 \]

Simplifying the above gives

\begin{align*} -\frac {\sqrt {3}\, \arctan \left (\frac {\left (-3 x +y \left (x \right )+6\right ) \sqrt {3}}{6+3 y \left (x \right )+3 x}\right )}{3}-\frac {\ln \left (\frac {y \left (x \right )^{2}+3 x^{2}+6 y \left (x \right )-6 x +12}{\left (x +y \left (x \right )+2\right )^{2}}\right )}{2}+\ln \left (2\right )+\ln \left (\frac {-1+x}{x +y \left (x \right )+2}\right ) &= \ln \left (-1+x \right )+c_2 \\ \end{align*}

Solving for initial conditions the solution is

\begin{align*} -\frac {\sqrt {3}\, \arctan \left (\frac {\left (-3 x +y \left (x \right )+6\right ) \sqrt {3}}{6+3 y \left (x \right )+3 x}\right )}{3}-\frac {\ln \left (\frac {y \left (x \right )^{2}+3 x^{2}+6 y \left (x \right )-6 x +12}{\left (x +y \left (x \right )+2\right )^{2}}\right )}{2}+\ln \left (2\right )+\ln \left (\frac {-1+x}{x +y \left (x \right )+2}\right ) &= \ln \left (-1+x \right )+\frac {\sqrt {3}\, \pi }{18} \\ \end{align*}

Figure 2.27: Slope field \(3 x -y \left (x \right )-6+\left (x +y \left (x \right )+2\right ) y^{\prime }\left (x \right ) = 0\)

Figure 2.28: Slope field \(3 x -y \left (x \right )-6+\left (x +y \left (x \right )+2\right ) y^{\prime }\left (x \right ) = 0\)

Summary of solutions found

\begin{align*} -\frac {\sqrt {3}\, \arctan \left (\frac {\left (-3 x +y \left (x \right )+6\right ) \sqrt {3}}{6+3 y \left (x \right )+3 x}\right )}{3}-\frac {\ln \left (\frac {y \left (x \right )^{2}+3 x^{2}+6 y \left (x \right )-6 x +12}{\left (x +y \left (x \right )+2\right )^{2}}\right )}{2}+\ln \left (2\right )+\ln \left (\frac {x -1}{x +y \left (x \right )+2}\right ) &= \ln \left (x -1\right )+\frac {\sqrt {3}\, \pi }{18} \\ \end{align*}

2.1.9.4 Solved using first_order_ode_LIE

1.086 (sec)

Entering first order ode LIE solver

\begin{align*} 3 x -y \left (x \right )-6+\left (x +y \left (x \right )+2\right ) y^{\prime }\left (x \right )&=0 \\ y \left (2\right ) &= -2 \\ \end{align*}

Writing the ode as

\begin{align*} y^{\prime }\left (x \right )&=\frac {-3 x +y +6}{x +y +2}\\ y^{\prime }\left (x \right )&= \omega \left ( x,y \left (x \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y \left (x \right )}-\xi _{x}\right ) -\omega ^{2}\xi _{y \left (x \right )}-\omega _{x}\xi -\omega _{y \left (x \right )}\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}+\frac {\left (-3 x +y +6\right ) \left (b_{3}-a_{2}\right )}{x +y +2}-\frac {\left (-3 x +y +6\right )^{2} a_{3}}{\left (x +y +2\right )^{2}}-\left (-\frac {3}{x +y +2}-\frac {-3 x +y +6}{\left (x +y +2\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {1}{x +y +2}-\frac {-3 x +y +6}{\left (x +y +2\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \frac {3 x^{2} a_{2}-9 x^{2} a_{3}-3 x^{2} b_{2}-3 x^{2} b_{3}+6 x y a_{2}+6 x y a_{3}+2 x y b_{2}-6 x y b_{3}-y^{2} a_{2}+3 y^{2} a_{3}+y^{2} b_{2}+y^{2} b_{3}+12 x a_{2}+36 x a_{3}-4 x b_{1}+8 x b_{2}+4 y a_{1}-8 y a_{2}+4 y b_{2}+12 y b_{3}+12 a_{1}-12 a_{2}-36 a_{3}+4 b_{1}+4 b_{2}+12 b_{3}}{\left (x +y +2\right )^{2}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} 3 x^{2} a_{2}-9 x^{2} a_{3}-3 x^{2} b_{2}-3 x^{2} b_{3}+6 x y a_{2}+6 x y a_{3}+2 x y b_{2}-6 x y b_{3}-y^{2} a_{2}+3 y^{2} a_{3}+y^{2} b_{2}+y^{2} b_{3}+12 x a_{2}+36 x a_{3}-4 x b_{1}+8 x b_{2}+4 y a_{1}-8 y a_{2}+4 y b_{2}+12 y b_{3}+12 a_{1}-12 a_{2}-36 a_{3}+4 b_{1}+4 b_{2}+12 b_{3} = 0 \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} 3 a_{2} v_{1}^{2}+6 a_{2} v_{1} v_{2}-a_{2} v_{2}^{2}-9 a_{3} v_{1}^{2}+6 a_{3} v_{1} v_{2}+3 a_{3} v_{2}^{2}-3 b_{2} v_{1}^{2}+2 b_{2} v_{1} v_{2}+b_{2} v_{2}^{2}-3 b_{3} v_{1}^{2}-6 b_{3} v_{1} v_{2}+b_{3} v_{2}^{2}+4 a_{1} v_{2}+12 a_{2} v_{1}-8 a_{2} v_{2}+36 a_{3} v_{1}-4 b_{1} v_{1}+8 b_{2} v_{1}+4 b_{2} v_{2}+12 b_{3} v_{2}+12 a_{1}-12 a_{2}-36 a_{3}+4 b_{1}+4 b_{2}+12 b_{3} = 0 \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} \left (3 a_{2}-9 a_{3}-3 b_{2}-3 b_{3}\right ) v_{1}^{2}+\left (6 a_{2}+6 a_{3}+2 b_{2}-6 b_{3}\right ) v_{1} v_{2}+\left (12 a_{2}+36 a_{3}-4 b_{1}+8 b_{2}\right ) v_{1}+\left (-a_{2}+3 a_{3}+b_{2}+b_{3}\right ) v_{2}^{2}+\left (4 a_{1}-8 a_{2}+4 b_{2}+12 b_{3}\right ) v_{2}+12 a_{1}-12 a_{2}-36 a_{3}+4 b_{1}+4 b_{2}+12 b_{3} = 0 \end{equation}

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

\begin{align*} 4 a_{1}-8 a_{2}+4 b_{2}+12 b_{3}&=0\\ -a_{2}+3 a_{3}+b_{2}+b_{3}&=0\\ 3 a_{2}-9 a_{3}-3 b_{2}-3 b_{3}&=0\\ 6 a_{2}+6 a_{3}+2 b_{2}-6 b_{3}&=0\\ 12 a_{2}+36 a_{3}-4 b_{1}+8 b_{2}&=0\\ 12 a_{1}-12 a_{2}-36 a_{3}+4 b_{1}+4 b_{2}+12 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

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

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= y +3 - \left (\frac {-3 x +y +6}{x +y +2}\right ) \left (x -1\right ) \\ &= \frac {3 x^{2}+y^{2}-6 x +6 y +12}{x +y +2}\\ \xi &= 0 \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 \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {3 x^{2}+y^{2}-6 x +6 y +12}{x +y +2}}} dy \end{align*}

Which results in

\begin{align*} S&= \frac {\ln \left (3 x^{2}+y^{2}-6 x +6 y +12\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 y +6\right ) \sqrt {3}}{-6+6 x}\right )}{3} \end{align*}

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) &= \frac {-3 x +y +6}{x +y +2} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {3 x -y -6}{3 x^{2}+y^{2}-6 x +6 y +12}\\ S_{y} &= \frac {x +y +2}{3 x^{2}+y^{2}-6 x +6 y +12} \end{align*}

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

\begin{align*} \frac {dS}{dR} &= 0\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} &= 0 \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 {0\, dR} + c_2 \\ S \left (R \right ) &= 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 {\ln \left (y \left (x \right )^{2}+3 x^{2}+6 y \left (x \right )-6 x +12\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (y \left (x \right )+3\right ) \sqrt {3}}{-3+3 x}\right )}{3} = c_2 \end{align*}

The following diagram shows solution curves of the original ode and how they transform in the canonical coordinates space using the mapping shown.

Original ode in \(x,y\) coordinates	Canonical coordinates transformation	ODE in canonical coordinates \((R,S)\)
\( \frac {dy}{dx} = \frac {-3 x +y +6}{x +y +2}\)		\( \frac {d S}{d R} = 0\)
	\(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (3 x^{2}+y^{2}-6 x +6 y +12\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (y +3\right ) \sqrt {3}}{-3+3 x}\right )}{3} \end {aligned} \)

Solving for initial conditions the solution is

\begin{align*} \frac {\ln \left (y \left (x \right )^{2}+3 x^{2}+6 y \left (x \right )-6 x +12\right )}{2}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (y \left (x \right )+3\right ) \sqrt {3}}{-3+3 x}\right )}{3} &= \frac {\sqrt {3}\, \pi }{18}+\ln \left (2\right ) \\ \end{align*}

Solving for \(y \left (x \right )\) gives

\begin{align*} y \left (x \right ) &= \tan \left (\operatorname {RootOf}\left (6 \sqrt {3}\, \ln \left (2\right )-3 \sqrt {3}\, \ln \left (3 x^{2} \tan \left (\textit {\_Z} \right )^{2}-6 x \tan \left (\textit {\_Z} \right )^{2}+3 x^{2}+3 \tan \left (\textit {\_Z} \right )^{2}-6 x +3\right )+\pi -6 \textit {\_Z} \right )\right ) \sqrt {3}\, x -\tan \left (\operatorname {RootOf}\left (6 \sqrt {3}\, \ln \left (2\right )-3 \sqrt {3}\, \ln \left (3 x^{2} \tan \left (\textit {\_Z} \right )^{2}-6 x \tan \left (\textit {\_Z} \right )^{2}+3 x^{2}+3 \tan \left (\textit {\_Z} \right )^{2}-6 x +3\right )+\pi -6 \textit {\_Z} \right )\right ) \sqrt {3}-3 \\ \end{align*}

Figure 2.29: Slope field \(3 x -y \left (x \right )-6+\left (x +y \left (x \right )+2\right ) y^{\prime }\left (x \right ) = 0\)

Summary of solutions found

\begin{align*} y \left (x \right ) &= \tan \left (\operatorname {RootOf}\left (6 \sqrt {3}\, \ln \left (2\right )-3 \sqrt {3}\, \ln \left (3 x^{2} \tan \left (\textit {\_Z} \right )^{2}-6 x \tan \left (\textit {\_Z} \right )^{2}+3 x^{2}+3 \tan \left (\textit {\_Z} \right )^{2}-6 x +3\right )+\pi -6 \textit {\_Z} \right )\right ) \sqrt {3}\, x -\tan \left (\operatorname {RootOf}\left (6 \sqrt {3}\, \ln \left (2\right )-3 \sqrt {3}\, \ln \left (3 x^{2} \tan \left (\textit {\_Z} \right )^{2}-6 x \tan \left (\textit {\_Z} \right )^{2}+3 x^{2}+3 \tan \left (\textit {\_Z} \right )^{2}-6 x +3\right )+\pi -6 \textit {\_Z} \right )\right ) \sqrt {3}-3 \\ \end{align*}

2.1.9.5 ✓ Maple. Time used: 1.166 (sec). Leaf size: 51

ode:=3*x-y(x)-6+(x+y(x)+2)*diff(y(x),x) = 0; 
ic:=[y(2) = -2]; 
dsolve([ode,op(ic)],y(x), singsol=all);
 

\[ y = -3-\tan \left (\operatorname {RootOf}\left (-3 \sqrt {3}\, \ln \left (\left (x -1\right )^{2} \sec \left (\textit {\_Z} \right )^{2}\right )-3 \sqrt {3}\, \ln \left (3\right )+6 \sqrt {3}\, \ln \left (2\right )+\pi +6 \textit {\_Z} \right )\right ) \sqrt {3}\, \left (x -1\right ) \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying homogeneous C 
trying homogeneous types: 
trying homogeneous D 
<- homogeneous successful 
<- homogeneous successful
 

2.1.9.6 ✓ Mathematica. Time used: 0.085 (sec). Leaf size: 90

ode=(3*x-y[x]-6)+(x+y[x]+2)*D[y[x],x]==0; 
ic=y[2]==-2; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\frac {\arctan \left (\frac {-y(x)+3 x-6}{\sqrt {3} (y(x)+x+2)}\right )}{\sqrt {3}}+\log (2)=\frac {1}{2} \log \left (\frac {3 x^2+y(x)^2+6 y(x)-6 x+12}{(x-1)^2}\right )+\log (x-1)+\frac {1}{18} \left (\sqrt {3} \pi +18 \log (2)-9 \log (4)\right ),y(x)\right ] \]

2.1.9.7 ✓ Sympy. Time used: 3.961 (sec). Leaf size: 58

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(3*x + (x + y(x) + 2)*Derivative(y(x), x) - y(x) - 6,0) 
ics = {y(2): -2} 
dsolve(ode,func=y(x),ics=ics)
 

\[ \log {\left (x - 1 \right )} = - \log {\left (\sqrt {3 + \frac {\left (y{\left (x \right )} + 3\right )^{2}}{\left (x - 1\right )^{2}}} \right )} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3} \left (y{\left (x \right )} + 3\right )}{3 \left (x - 1\right )} \right )}}{3} + \frac {\sqrt {3} \pi }{18} + \log {\left (2 \right )} \]