1.285 problem 286

Internal problem ID [8622]
Internal file name [OUTPUT/7555_Sunday_June_05_2022_11_05_28_PM_32476382/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 286.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_calculated"

Maple gives the following as the ode type

[[_homogeneous, `class C`], _rational]

\[ \boxed {\left (2 y-3 x +1\right )^{2} y^{\prime }-\left (3 y-2 x -4\right )^{2}=0} \]

1.285.1 Solving as homogeneousTypeMapleC ode

Let \(Y = y +y_{0}\) and \(X = x +x_{0}\) then the above is transformed to new ode in \(Y(X)\) \begin {align*} \frac {d}{d X}Y \left (X \right ) = \frac {\left (3 Y \left (X \right )+3 y_{0} -2 x_{0} -2 X -4\right )^{2}}{\left (2 Y \left (X \right )+2 y_{0} -3 x_{0} -3 X +1\right )^{2}} \end {align*}

Solving for possible values of \(x_{0}\) and \(y_{0}\) which makes the above ode a homogeneous ode results in \begin {align*} x_{0}&={\frac {11}{5}}\\ y_{0}&={\frac {14}{5}} \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 {4 X^{2}-12 Y \left (X \right ) X +9 Y \left (X \right )^{2}}{9 X^{2}-12 Y \left (X \right ) X +4 Y \left (X \right )^{2}} \end {align*}

In canonical form, the ODE is \begin {align*} Y' &= F(X,Y)\\ &= \frac {4 X^{2}-12 Y X +9 Y^{2}}{9 X^{2}-12 Y X +4 Y^{2}}\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 \(M=4 X^{2}-12 Y X +9 Y^{2}\) and \(N=9 X^{2}-12 Y X +4 Y^{2}\) are both homogeneous and of the same order \(n=2\). 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 {\left (3 u -2\right )^{2}}{\left (2 u -3\right )^{2}}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {\left (3 u \left (X \right )-2\right )^{2}}{\left (2 u \left (X \right )-3\right )^{2}}-u \left (X \right )}{X} \end {align*}

Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {\left (3 u \left (X \right )-2\right )^{2}}{\left (2 u \left (X \right )-3\right )^{2}}-u \left (X \right )}{X} = 0 \] Or \[ 4 \left (\frac {d}{d X}u \left (X \right )\right ) u \left (X \right )^{2} X -12 \left (\frac {d}{d X}u \left (X \right )\right ) u \left (X \right ) X +4 u \left (X \right )^{3}+9 \left (\frac {d}{d X}u \left (X \right )\right ) X -21 u \left (X \right )^{2}+21 u \left (X \right )-4 = 0 \] Or \[ 4 X \left (u \left (X \right )-\frac {3}{2}\right )^{2} \left (\frac {d}{d X}u \left (X \right )\right )+4 u \left (X \right )^{3}-21 u \left (X \right )^{2}+21 u \left (X \right )-4 = 0 \] Which is now solved as separable in \(u \left (X \right )\). Which is now solved in \(u \left (X \right )\). In canonical form the ODE is \begin {align*} u' &= F(X,u)\\ &= f( X) g(u)\\ &= -\frac {4 u^{3}-21 u^{2}+21 u -4}{4 X \left (u -\frac {3}{2}\right )^{2}} \end {align*}

Where \(f(X)=-\frac {1}{X}\) and \(g(u)=\frac {4 u^{3}-21 u^{2}+21 u -4}{4 \left (u -\frac {3}{2}\right )^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {4 u^{3}-21 u^{2}+21 u -4}{4 \left (u -\frac {3}{2}\right )^{2}}} \,du &= -\frac {1}{X} \,d X \\ \int { \frac {1}{\frac {4 u^{3}-21 u^{2}+21 u -4}{4 \left (u -\frac {3}{2}\right )^{2}}} \,du} &= \int {-\frac {1}{X} \,d X} \\ \frac {5 \ln \left (u -4\right )}{9}-\frac {\ln \left (u -1\right )}{9}+\frac {5 \ln \left (u -\frac {1}{4}\right )}{9}&=-\ln \left (X \right )+c_{2} \\ \end{align*} The above can be written as \begin {align*} \frac {5 \ln \left (u -4\right )-\ln \left (u -1\right )+5 \ln \left (u -\frac {1}{4}\right )}{9} &= -\ln \left (X \right )+c_{2}\\ 5 \ln \left (u -4\right )-\ln \left (u -1\right )+5 \ln \left (u -\frac {1}{4}\right ) &= \left (9\right ) \left (-\ln \left (X \right )+c_{2}\right ) \\ &= -9 \ln \left (X \right )+9 c_{2} \end {align*}

Raising both side to exponential gives \begin {align*} {\mathrm e}^{5 \ln \left (u -4\right )-\ln \left (u -1\right )+5 \ln \left (u -\frac {1}{4}\right )} &= {\mathrm e}^{-9 \ln \left (X \right )+9 c_{2}} \end {align*}

Which simplifies to \begin {align*} \frac {\left (u -4\right )^{5} \left (4 u -1\right )^{5}}{1024 u -1024} &= \frac {9 c_{2}}{X^{9}}\\ &= \frac {c_{3}}{X^{9}} \end {align*}

Which simplifies to \[ u \left (X \right ) = \operatorname {RootOf}\left (1024 \textit {\_Z}^{10}-21760 \textit {\_Z}^{9}+190080 \textit {\_Z}^{8}-873120 \textit {\_Z}^{7}+2235540 \textit {\_Z}^{6}-3122577 \textit {\_Z}^{5}+2235540 \textit {\_Z}^{4}-873120 \textit {\_Z}^{3}+190080 \textit {\_Z}^{2}+\left (-\frac {1024 c_{3} {\mathrm e}^{9 c_{2}}}{X^{9}}-21760\right ) \textit {\_Z} +\frac {1024 c_{3} {\mathrm e}^{9 c_{2}}}{X^{9}}+1024\right ) \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ Y \left (X \right ) = X \operatorname {RootOf}\left (1024 \textit {\_Z}^{10} X^{9}-21760 \textit {\_Z}^{9} X^{9}+190080 \textit {\_Z}^{8} X^{9}-873120 \textit {\_Z}^{7} X^{9}+2235540 \textit {\_Z}^{6} X^{9}-3122577 \textit {\_Z}^{5} X^{9}+2235540 \textit {\_Z}^{4} X^{9}-873120 \textit {\_Z}^{3} X^{9}+190080 \textit {\_Z}^{2} X^{9}+\left (-1024 c_{3} {\mathrm e}^{9 c_{2}}-21760 X^{9}\right ) \textit {\_Z} +1024 c_{3} {\mathrm e}^{9 c_{2}}+1024 X^{9}\right ) \] Using the solution for \(Y(X)\) \begin {align*} Y \left (X \right ) = X \operatorname {RootOf}\left (1024 \textit {\_Z}^{10} X^{9}-21760 \textit {\_Z}^{9} X^{9}+190080 \textit {\_Z}^{8} X^{9}-873120 \textit {\_Z}^{7} X^{9}+2235540 \textit {\_Z}^{6} X^{9}-3122577 \textit {\_Z}^{5} X^{9}+2235540 \textit {\_Z}^{4} X^{9}-873120 \textit {\_Z}^{3} X^{9}+190080 \textit {\_Z}^{2} X^{9}+\left (-1024 c_{3} {\mathrm e}^{9 c_{2}}-21760 X^{9}\right ) \textit {\_Z} +1024 c_{3} {\mathrm e}^{9 c_{2}}+1024 X^{9}\right ) \end {align*}

And replacing back terms in the above solution using \begin {align*} Y &= y +y_{0}\\ X &= x +x_{0} \end {align*}

Or \begin {align*} Y &= y +\frac {14}{5}\\ X &= x +\frac {11}{5} \end {align*}

Then the solution in \(y\) becomes \begin {align*} y-\frac {14}{5} = \left (x -\frac {11}{5}\right ) \operatorname {RootOf}\left (\left (2000000000 x^{9}-39600000000 x^{8}[phone] x^{7}-1788864000000 x^{6}[phone] x^{5}-12987152640000 x^{4}[phone] x^{3}-17959376793600 x^{2}[phone] x -2414538435584\right ) \textit {\_Z}^{10}+\left (-42500000000 x^{9}[phone] x^{8}-7405200000000 x^{7}[phone] x^{6}-125444088000000 x^{5}[phone] x^{4}-404766257280000 x^{3}[phone] x^{2}-209900216275200 x [phone]\right ) \textit {\_Z}^{9}+\left (371250000000 x^{9}-7350750000000 x^{8}[phone] x^{7}-332057880000000 x^{6}+1095791004000000 x^{5}-2410740208800000 x^{4}+3535752306240000 x^{3}-3333709317312000 x^{2}+1833540124521600 x -448198697105280\right ) \textit {\_Z}^{8}+\left (-1705312500000 x^{9}[phone] x^{8}-297133650000000 x^{7}+1525286070000000 x^{6}-5033444031000000 x^{5}+11073576868200000 x^{4}-16241246073360000 x^{3}+15313174869168000 x^{2}-8422246178042400 x +2058771287965920\right ) \textit {\_Z}^{7}+\left (4366289062500 x^{9}-86452523437500 x^{8}[phone] x^{7}-3905348658750000 x^{6}+12887650573875000 x^{5}-28352831262525000 x^{4}+41584152518370000 x^{3}-39207915231606000 x^{2}+21564353377383300 x -5271286381138140\right ) \textit {\_Z}^{6}+\left (-6098783203125 x^{9}[phone] x^{8}-1062651985312500 x^{7}+5454946857937500 x^{6}-18001324631193750 x^{5}+39602914188626250 x^{4}-58084274143318500 x^{3}+54765172763700300 x^{2}-30120845020035165 x +7362873227119707\right ) \textit {\_Z}^{5}+\left (4366289062500 x^{9}-86452523437500 x^{8}[phone] x^{7}-3905348658750000 x^{6}+12887650573875000 x^{5}-28352831262525000 x^{4}+41584152518370000 x^{3}-39207915231606000 x^{2}+21564353377383300 x -5271286381138140\right ) \textit {\_Z}^{4}+\left (-1705312500000 x^{9}[phone] x^{8}-297133650000000 x^{7}+1525286070000000 x^{6}-5033444031000000 x^{5}+11073576868200000 x^{4}-16241246073360000 x^{3}+15313174869168000 x^{2}-8422246178042400 x +2058771287965920\right ) \textit {\_Z}^{3}+\left (371250000000 x^{9}-7350750000000 x^{8}[phone] x^{7}-332057880000000 x^{6}+1095791004000000 x^{5}-2410740208800000 x^{4}+3535752306240000 x^{3}-3333709317312000 x^{2}+1833540124521600 x -448198697105280\right ) \textit {\_Z}^{2}+\left (-2000000000 c_{3} {\mathrm e}^{9 c_{2}}-42500000000 x^{9}[phone] x^{8}-7405200000000 x^{7}[phone] x^{6}-125444088000000 x^{5}[phone] x^{4}-404766257280000 x^{3}[phone] x^{2}-209900216275200 x [phone]\right ) \textit {\_Z} [phone] c_{3} {\mathrm e}^{9 c_{2}}[phone] x^{9}-39600000000 x^{8}[phone] x^{7}-1788864000000 x^{6}[phone] x^{5}-12987152640000 x^{4}[phone] x^{3}-17959376793600 x^{2}[phone] x -2414538435584\right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y-\frac {14}{5} &= \left (x -\frac {11}{5}\right ) \operatorname {RootOf}\left (\left (2000000000 x^{9}-39600000000 x^{8}[phone] x^{7}-1788864000000 x^{6}[phone] x^{5}-12987152640000 x^{4}[phone] x^{3}-17959376793600 x^{2}[phone] x -2414538435584\right ) \textit {\_Z}^{10}+\left (-42500000000 x^{9}[phone] x^{8}-7405200000000 x^{7}[phone] x^{6}-125444088000000 x^{5}[phone] x^{4}-404766257280000 x^{3}[phone] x^{2}-209900216275200 x [phone]\right ) \textit {\_Z}^{9}+\left (371250000000 x^{9}-7350750000000 x^{8}[phone] x^{7}-332057880000000 x^{6}+1095791004000000 x^{5}-2410740208800000 x^{4}+3535752306240000 x^{3}-3333709317312000 x^{2}+1833540124521600 x -448198697105280\right ) \textit {\_Z}^{8}+\left (-1705312500000 x^{9}[phone] x^{8}-297133650000000 x^{7}+1525286070000000 x^{6}-5033444031000000 x^{5}+11073576868200000 x^{4}-16241246073360000 x^{3}+15313174869168000 x^{2}-8422246178042400 x +2058771287965920\right ) \textit {\_Z}^{7}+\left (4366289062500 x^{9}-86452523437500 x^{8}[phone] x^{7}-3905348658750000 x^{6}+12887650573875000 x^{5}-28352831262525000 x^{4}+41584152518370000 x^{3}-39207915231606000 x^{2}+21564353377383300 x -5271286381138140\right ) \textit {\_Z}^{6}+\left (-6098783203125 x^{9}[phone] x^{8}-1062651985312500 x^{7}+5454946857937500 x^{6}-18001324631193750 x^{5}+39602914188626250 x^{4}-58084274143318500 x^{3}+54765172763700300 x^{2}-30120845020035165 x +7362873227119707\right ) \textit {\_Z}^{5}+\left (4366289062500 x^{9}-86452523437500 x^{8}[phone] x^{7}-3905348658750000 x^{6}+12887650573875000 x^{5}-28352831262525000 x^{4}+41584152518370000 x^{3}-39207915231606000 x^{2}+21564353377383300 x -5271286381138140\right ) \textit {\_Z}^{4}+\left (-1705312500000 x^{9}[phone] x^{8}-297133650000000 x^{7}+1525286070000000 x^{6}-5033444031000000 x^{5}+11073576868200000 x^{4}-16241246073360000 x^{3}+15313174869168000 x^{2}-8422246178042400 x +2058771287965920\right ) \textit {\_Z}^{3}+\left (371250000000 x^{9}-7350750000000 x^{8}[phone] x^{7}-332057880000000 x^{6}+1095791004000000 x^{5}-2410740208800000 x^{4}+3535752306240000 x^{3}-3333709317312000 x^{2}+1833540124521600 x -448198697105280\right ) \textit {\_Z}^{2}+\left (-2000000000 c_{3} {\mathrm e}^{9 c_{2}}-42500000000 x^{9}[phone] x^{8}-7405200000000 x^{7}[phone] x^{6}-125444088000000 x^{5}[phone] x^{4}-404766257280000 x^{3}[phone] x^{2}-209900216275200 x [phone]\right ) \textit {\_Z} [phone] c_{3} {\mathrm e}^{9 c_{2}}[phone] x^{9}-39600000000 x^{8}[phone] x^{7}-1788864000000 x^{6}[phone] x^{5}-12987152640000 x^{4}[phone] x^{3}-17959376793600 x^{2}[phone] x -2414538435584\right ) \\ \end{align*}

Figure 343: Slope field plot

Verification of solutions

\[ y-\frac {14}{5} = \left (x -\frac {11}{5}\right ) \operatorname {RootOf}\left (\left (2000000000 x^{9}-39600000000 x^{8}[phone] x^{7}-1788864000000 x^{6}[phone] x^{5}-12987152640000 x^{4}[phone] x^{3}-17959376793600 x^{2}[phone] x -2414538435584\right ) \textit {\_Z}^{10}+\left (-42500000000 x^{9}[phone] x^{8}-7405200000000 x^{7}[phone] x^{6}-125444088000000 x^{5}[phone] x^{4}-404766257280000 x^{3}[phone] x^{2}-209900216275200 x [phone]\right ) \textit {\_Z}^{9}+\left (371250000000 x^{9}-7350750000000 x^{8}[phone] x^{7}-332057880000000 x^{6}+1095791004000000 x^{5}-2410740208800000 x^{4}+3535752306240000 x^{3}-3333709317312000 x^{2}+1833540124521600 x -448198697105280\right ) \textit {\_Z}^{8}+\left (-1705312500000 x^{9}[phone] x^{8}-297133650000000 x^{7}+1525286070000000 x^{6}-5033444031000000 x^{5}+11073576868200000 x^{4}-16241246073360000 x^{3}+15313174869168000 x^{2}-8422246178042400 x +2058771287965920\right ) \textit {\_Z}^{7}+\left (4366289062500 x^{9}-86452523437500 x^{8}[phone] x^{7}-3905348658750000 x^{6}+12887650573875000 x^{5}-28352831262525000 x^{4}+41584152518370000 x^{3}-39207915231606000 x^{2}+21564353377383300 x -5271286381138140\right ) \textit {\_Z}^{6}+\left (-6098783203125 x^{9}[phone] x^{8}-1062651985312500 x^{7}+5454946857937500 x^{6}-18001324631193750 x^{5}+39602914188626250 x^{4}-58084274143318500 x^{3}+54765172763700300 x^{2}-30120845020035165 x +7362873227119707\right ) \textit {\_Z}^{5}+\left (4366289062500 x^{9}-86452523437500 x^{8}[phone] x^{7}-3905348658750000 x^{6}+12887650573875000 x^{5}-28352831262525000 x^{4}+41584152518370000 x^{3}-39207915231606000 x^{2}+21564353377383300 x -5271286381138140\right ) \textit {\_Z}^{4}+\left (-1705312500000 x^{9}[phone] x^{8}-297133650000000 x^{7}+1525286070000000 x^{6}-5033444031000000 x^{5}+11073576868200000 x^{4}-16241246073360000 x^{3}+15313174869168000 x^{2}-8422246178042400 x +2058771287965920\right ) \textit {\_Z}^{3}+\left (371250000000 x^{9}-7350750000000 x^{8}[phone] x^{7}-332057880000000 x^{6}+1095791004000000 x^{5}-2410740208800000 x^{4}+3535752306240000 x^{3}-3333709317312000 x^{2}+1833540124521600 x -448198697105280\right ) \textit {\_Z}^{2}+\left (-2000000000 c_{3} {\mathrm e}^{9 c_{2}}-42500000000 x^{9}[phone] x^{8}-7405200000000 x^{7}[phone] x^{6}-125444088000000 x^{5}[phone] x^{4}-404766257280000 x^{3}[phone] x^{2}-209900216275200 x [phone]\right ) \textit {\_Z} [phone] c_{3} {\mathrm e}^{9 c_{2}}[phone] x^{9}-39600000000 x^{8}[phone] x^{7}-1788864000000 x^{6}[phone] x^{5}-12987152640000 x^{4}[phone] x^{3}-17959376793600 x^{2}[phone] x -2414538435584\right ) \] Verified OK.

1.285.2 Solving as first order ode lie symmetry calculated ode

Writing the ode as \begin {align*} y^{\prime }&=\frac {4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16}{9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1}\\ 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*}

The type of this ode is not in the lookup table. 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 (4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16\right ) \left (b_{3}-a_{2}\right )}{9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1}-\frac {\left (4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16\right )^{2} a_{3}}{\left (9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1\right )^{2}}-\left (\frac {8 x -12 y +16}{9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1}-\frac {\left (4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16\right ) \left (18 x -12 y -6\right )}{\left (9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {18 y -12 x -24}{9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1}-\frac {\left (4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16\right ) \left (-12 x +8 y +4\right )}{\left (9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {36 x^{4} a_{2}+16 x^{4} a_{3}-141 x^{4} b_{2}-36 x^{4} b_{3}-96 x^{3} y a_{2}-96 x^{3} y a_{3}+346 x^{3} y b_{2}+96 x^{3} y b_{3}+111 x^{2} y^{2} a_{2}+276 x^{2} y^{2} a_{3}-276 x^{2} y^{2} b_{2}-111 x^{2} y^{2} b_{3}-96 x \,y^{3} a_{2}-346 x \,y^{3} a_{3}+96 x \,y^{3} b_{2}+96 x \,y^{3} b_{3}+36 y^{4} a_{2}+141 y^{4} a_{3}-16 y^{4} b_{2}-36 y^{4} b_{3}-48 x^{3} a_{2}+128 x^{3} a_{3}-60 x^{3} b_{1}+140 x^{3} b_{2}-120 x^{3} b_{3}+60 x^{2} y a_{1}+144 x^{2} y a_{2}-744 x^{2} y a_{3}+130 x^{2} y b_{1}-452 x^{2} y b_{2}+352 x^{2} y b_{3}-130 x \,y^{2} a_{1}+32 x \,y^{2} a_{2}+1328 x \,y^{2} a_{3}-60 x \,y^{2} b_{1}+276 x \,y^{2} b_{2}-486 x \,y^{2} b_{3}+60 y^{3} a_{1}-60 y^{3} a_{2}-650 y^{3} a_{3}-32 y^{3} b_{2}+192 y^{3} b_{3}-168 x^{2} a_{1}-228 x^{2} a_{2}+384 x^{2} a_{3}+32 x^{2} b_{1}+206 x^{2} b_{2}-52 x^{2} b_{3}+464 x y a_{1}+104 x y a_{2}-1432 x y a_{3}-236 x y b_{1}-38 x y b_{2}+256 x y b_{3}-218 y^{2} a_{1}-23 y^{2} a_{2}+964 y^{2} a_{3}+132 y^{2} b_{1}-24 y^{2} b_{2}-87 y^{2} b_{3}-280 x a_{1}+32 x a_{2}+512 x a_{3}+260 x b_{1}-76 x b_{2}+80 x b_{3}+100 y a_{1}+40 y a_{2}-656 y a_{3}-110 y b_{1}-8 y b_{2}-128 y b_{3}+112 a_{1}+16 a_{2}+256 a_{3}-88 b_{1}-b_{2}-16 b_{3}}{\left (9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -36 x^{4} a_{2}-16 x^{4} a_{3}+141 x^{4} b_{2}+36 x^{4} b_{3}+96 x^{3} y a_{2}+96 x^{3} y a_{3}-346 x^{3} y b_{2}-96 x^{3} y b_{3}-111 x^{2} y^{2} a_{2}-276 x^{2} y^{2} a_{3}+276 x^{2} y^{2} b_{2}+111 x^{2} y^{2} b_{3}+96 x \,y^{3} a_{2}+346 x \,y^{3} a_{3}-96 x \,y^{3} b_{2}-96 x \,y^{3} b_{3}-36 y^{4} a_{2}-141 y^{4} a_{3}+16 y^{4} b_{2}+36 y^{4} b_{3}+48 x^{3} a_{2}-128 x^{3} a_{3}+60 x^{3} b_{1}-140 x^{3} b_{2}+120 x^{3} b_{3}-60 x^{2} y a_{1}-144 x^{2} y a_{2}+744 x^{2} y a_{3}-130 x^{2} y b_{1}+452 x^{2} y b_{2}-352 x^{2} y b_{3}+130 x \,y^{2} a_{1}-32 x \,y^{2} a_{2}-1328 x \,y^{2} a_{3}+60 x \,y^{2} b_{1}-276 x \,y^{2} b_{2}+486 x \,y^{2} b_{3}-60 y^{3} a_{1}+60 y^{3} a_{2}+650 y^{3} a_{3}+32 y^{3} b_{2}-192 y^{3} b_{3}+168 x^{2} a_{1}+228 x^{2} a_{2}-384 x^{2} a_{3}-32 x^{2} b_{1}-206 x^{2} b_{2}+52 x^{2} b_{3}-464 x y a_{1}-104 x y a_{2}+1432 x y a_{3}+236 x y b_{1}+38 x y b_{2}-256 x y b_{3}+218 y^{2} a_{1}+23 y^{2} a_{2}-964 y^{2} a_{3}-132 y^{2} b_{1}+24 y^{2} b_{2}+87 y^{2} b_{3}+280 x a_{1}-32 x a_{2}-512 x a_{3}-260 x b_{1}+76 x b_{2}-80 x b_{3}-100 y a_{1}-40 y a_{2}+656 y a_{3}+110 y b_{1}+8 y b_{2}+128 y b_{3}-112 a_{1}-16 a_{2}-256 a_{3}+88 b_{1}+b_{2}+16 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} -36 a_{2} v_{1}^{4}+96 a_{2} v_{1}^{3} v_{2}-111 a_{2} v_{1}^{2} v_{2}^{2}+96 a_{2} v_{1} v_{2}^{3}-36 a_{2} v_{2}^{4}-16 a_{3} v_{1}^{4}+96 a_{3} v_{1}^{3} v_{2}-276 a_{3} v_{1}^{2} v_{2}^{2}+346 a_{3} v_{1} v_{2}^{3}-141 a_{3} v_{2}^{4}+141 b_{2} v_{1}^{4}-346 b_{2} v_{1}^{3} v_{2}+276 b_{2} v_{1}^{2} v_{2}^{2}-96 b_{2} v_{1} v_{2}^{3}+16 b_{2} v_{2}^{4}+36 b_{3} v_{1}^{4}-96 b_{3} v_{1}^{3} v_{2}+111 b_{3} v_{1}^{2} v_{2}^{2}-96 b_{3} v_{1} v_{2}^{3}+36 b_{3} v_{2}^{4}-60 a_{1} v_{1}^{2} v_{2}+130 a_{1} v_{1} v_{2}^{2}-60 a_{1} v_{2}^{3}+48 a_{2} v_{1}^{3}-144 a_{2} v_{1}^{2} v_{2}-32 a_{2} v_{1} v_{2}^{2}+60 a_{2} v_{2}^{3}-128 a_{3} v_{1}^{3}+744 a_{3} v_{1}^{2} v_{2}-1328 a_{3} v_{1} v_{2}^{2}+650 a_{3} v_{2}^{3}+60 b_{1} v_{1}^{3}-130 b_{1} v_{1}^{2} v_{2}+60 b_{1} v_{1} v_{2}^{2}-140 b_{2} v_{1}^{3}+452 b_{2} v_{1}^{2} v_{2}-276 b_{2} v_{1} v_{2}^{2}+32 b_{2} v_{2}^{3}+120 b_{3} v_{1}^{3}-352 b_{3} v_{1}^{2} v_{2}+486 b_{3} v_{1} v_{2}^{2}-192 b_{3} v_{2}^{3}+168 a_{1} v_{1}^{2}-464 a_{1} v_{1} v_{2}+218 a_{1} v_{2}^{2}+228 a_{2} v_{1}^{2}-104 a_{2} v_{1} v_{2}+23 a_{2} v_{2}^{2}-384 a_{3} v_{1}^{2}+1432 a_{3} v_{1} v_{2}-964 a_{3} v_{2}^{2}-32 b_{1} v_{1}^{2}+236 b_{1} v_{1} v_{2}-132 b_{1} v_{2}^{2}-206 b_{2} v_{1}^{2}+38 b_{2} v_{1} v_{2}+24 b_{2} v_{2}^{2}+52 b_{3} v_{1}^{2}-256 b_{3} v_{1} v_{2}+87 b_{3} v_{2}^{2}+280 a_{1} v_{1}-100 a_{1} v_{2}-32 a_{2} v_{1}-40 a_{2} v_{2}-512 a_{3} v_{1}+656 a_{3} v_{2}-260 b_{1} v_{1}+110 b_{1} v_{2}+76 b_{2} v_{1}+8 b_{2} v_{2}-80 b_{3} v_{1}+128 b_{3} v_{2}-112 a_{1}-16 a_{2}-256 a_{3}+88 b_{1}+b_{2}+16 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 (-36 a_{2}-16 a_{3}+141 b_{2}+36 b_{3}\right ) v_{1}^{4}+\left (96 a_{2}+96 a_{3}-346 b_{2}-96 b_{3}\right ) v_{1}^{3} v_{2}+\left (48 a_{2}-128 a_{3}+60 b_{1}-140 b_{2}+120 b_{3}\right ) v_{1}^{3}+\left (-111 a_{2}-276 a_{3}+276 b_{2}+111 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (-60 a_{1}-144 a_{2}+744 a_{3}-130 b_{1}+452 b_{2}-352 b_{3}\right ) v_{1}^{2} v_{2}+\left (168 a_{1}+228 a_{2}-384 a_{3}-32 b_{1}-206 b_{2}+52 b_{3}\right ) v_{1}^{2}+\left (96 a_{2}+346 a_{3}-96 b_{2}-96 b_{3}\right ) v_{1} v_{2}^{3}+\left (130 a_{1}-32 a_{2}-1328 a_{3}+60 b_{1}-276 b_{2}+486 b_{3}\right ) v_{1} v_{2}^{2}+\left (-464 a_{1}-104 a_{2}+1432 a_{3}+236 b_{1}+38 b_{2}-256 b_{3}\right ) v_{1} v_{2}+\left (280 a_{1}-32 a_{2}-512 a_{3}-260 b_{1}+76 b_{2}-80 b_{3}\right ) v_{1}+\left (-36 a_{2}-141 a_{3}+16 b_{2}+36 b_{3}\right ) v_{2}^{4}+\left (-60 a_{1}+60 a_{2}+650 a_{3}+32 b_{2}-192 b_{3}\right ) v_{2}^{3}+\left (218 a_{1}+23 a_{2}-964 a_{3}-132 b_{1}+24 b_{2}+87 b_{3}\right ) v_{2}^{2}+\left (-100 a_{1}-40 a_{2}+656 a_{3}+110 b_{1}+8 b_{2}+128 b_{3}\right ) v_{2}-112 a_{1}-16 a_{2}-256 a_{3}+88 b_{1}+b_{2}+16 b_{3} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -111 a_{2}-276 a_{3}+276 b_{2}+111 b_{3}&=0\\ -36 a_{2}-141 a_{3}+16 b_{2}+36 b_{3}&=0\\ -36 a_{2}-16 a_{3}+141 b_{2}+36 b_{3}&=0\\ 96 a_{2}+96 a_{3}-346 b_{2}-96 b_{3}&=0\\ 96 a_{2}+346 a_{3}-96 b_{2}-96 b_{3}&=0\\ -60 a_{1}+60 a_{2}+650 a_{3}+32 b_{2}-192 b_{3}&=0\\ 48 a_{2}-128 a_{3}+60 b_{1}-140 b_{2}+120 b_{3}&=0\\ -464 a_{1}-104 a_{2}+1432 a_{3}+236 b_{1}+38 b_{2}-256 b_{3}&=0\\ -112 a_{1}-16 a_{2}-256 a_{3}+88 b_{1}+b_{2}+16 b_{3}&=0\\ -100 a_{1}-40 a_{2}+656 a_{3}+110 b_{1}+8 b_{2}+128 b_{3}&=0\\ -60 a_{1}-144 a_{2}+744 a_{3}-130 b_{1}+452 b_{2}-352 b_{3}&=0\\ 130 a_{1}-32 a_{2}-1328 a_{3}+60 b_{1}-276 b_{2}+486 b_{3}&=0\\ 168 a_{1}+228 a_{2}-384 a_{3}-32 b_{1}-206 b_{2}+52 b_{3}&=0\\ 218 a_{1}+23 a_{2}-964 a_{3}-132 b_{1}+24 b_{2}+87 b_{3}&=0\\ 280 a_{1}-32 a_{2}-512 a_{3}-260 b_{1}+76 b_{2}-80 b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=-\frac {11 b_{3}}{5}\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=-\frac {14 b_{3}}{5}\\ 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 &= x -\frac {11}{5} \\ \eta &= y -\frac {14}{5} \\ \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 -\frac {14}{5} - \left (\frac {4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16}{9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1}\right ) \left (x -\frac {11}{5}\right ) \\ &= \frac {-20 x^{3}+105 x^{2} y -105 x \,y^{2}+20 y^{3}-162 x^{2}+126 y x +63 y^{2}+180 x -315 y +162}{45 x^{2}-60 y x +20 y^{2}-30 x +20 y +5}\\ \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 {-20 x^{3}+105 x^{2} y -105 x \,y^{2}+20 y^{3}-162 x^{2}+126 y x +63 y^{2}+180 x -315 y +162}{45 x^{2}-60 y x +20 y^{2}-30 x +20 y +5}}} dy \end {align*}

Which results in \begin {align*} S&= \frac {5 \ln \left (-4 x +6+y \right )}{9}+\frac {5 \ln \left (-x -9+4 y \right )}{9}-\frac {\ln \left (-5 x -3+5 y \right )}{9} \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 {4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16}{9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {20 \left (x -\frac {3 y}{2}+2\right )^{2}}{\left (-5 y +3+5 x \right ) \left (-y -6+4 x \right ) \left (-4 y +9+x \right )}\\ S_{y} &= -\frac {45 \left (-\frac {2 y}{3}+x -\frac {1}{3}\right )^{2}}{\left (-5 y +3+5 x \right ) \left (-y -6+4 x \right ) \left (-4 y +9+x \right )} \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 \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = c_{1}\tag {4} \end {align*}

To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} \frac {5 \ln \left (-4 x +6+y\right )}{9}+\frac {5 \ln \left (-x -9+4 y\right )}{9}-\frac {\ln \left (-5 x +5 y-3\right )}{9} = c_{1} \end {align*}

Which simplifies to \begin {align*} \frac {5 \ln \left (-4 x +6+y\right )}{9}+\frac {5 \ln \left (-x -9+4 y\right )}{9}-\frac {\ln \left (-5 x +5 y-3\right )}{9} = c_{1} \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 \((R,S)\)
\( \frac {dy}{dx} = \frac {4 x^{2}-12 y x +9 y^{2}+16 x -24 y +16}{9 x^{2}-12 y x +4 y^{2}-6 x +4 y +1}\)		\( \frac {d S}{d R} = 0\)
	\(\!\begin {aligned} R&= x\\ S&= \frac {5 \ln \left (-4 x +6+y \right )}{9}+\frac {5 \ln \left (-x -9+4 y \right )}{9}-\frac {\ln \left (-5 x -3+5 y \right )}{9} \end {aligned} \)

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {5 \ln \left (-4 x +6+y\right )}{9}+\frac {5 \ln \left (-x -9+4 y\right )}{9}-\frac {\ln \left (-5 x +5 y-3\right )}{9} &= c_{1} \\ \end{align*}

Figure 344: Slope field plot

Verification of solutions

\[ \frac {5 \ln \left (-4 x +6+y\right )}{9}+\frac {5 \ln \left (-x -9+4 y\right )}{9}-\frac {\ln \left (-5 x +5 y-3\right )}{9} = c_{1} \] Verified OK.

1.285.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 y-3 x +1\right )^{2} y^{\prime }-\left (3 y-2 x -4\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\left (3 y-2 x -4\right )^{2}}{\left (2 y-3 x +1\right )^{2}} \end {array} \]

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`
 

✓ Solution by Maple

Time used: 1.953 (sec). Leaf size: 1337

dsolve((2*y(x)-3*x+1)^2*diff(y(x),x)-(3*y(x)-2*x-4)^2=0,y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 60.198 (sec). Leaf size: 3501

DSolve[(2*y[x]-3*x+1)^2*y'[x]-(3*y[x]-2*x-4)^2==0,y[x],x,IncludeSingularSolutions -> True]
 

Too large to display