Internal problem ID [9143]
Internal file name [OUTPUT/8078_Monday_June_06_2022_01_40_49_AM_79529064/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 809.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind", "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class C`], _rational, _Abel]
\[ \boxed {y^{\prime }-\frac {-125+300 x -240 x^{2}+64 x^{3}-80 y^{2}+64 y^{2} x +64 y^{3}}{\left (4 x -5\right )^{3}}=0} \]
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 {-125+300 x_{0} +300 X -240 \left (x_{0} +X \right )^{2}+64 \left (x_{0} +X \right )^{3}-80 \left (Y \left (X \right )+y_{0} \right )^{2}+64 \left (Y \left (X \right )+y_{0} \right )^{2} \left (x_{0} +X \right )+64 \left (Y \left (X \right )+y_{0} \right )^{3}}{64 \left (x_{0} +X \right )^{3}-240 \left (x_{0} +X \right )^{2}+300 x_{0} +300 X -125} \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 {5}{4}}\\ y_{0}&=0 \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 {64 X^{3}+64 Y \left (X \right )^{2} X +64 Y \left (X \right )^{3}}{64 X^{3}} \end {align*}
In canonical form, the ODE is \begin {align*} Y' &= F(X,Y)\\ &= \frac {X^{3}+X \,Y^{2}+Y^{3}}{X^{3}}\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 \(M=X^{3}+X \,Y^{2}+Y^{3}\) and \(N=X^{3}\) are both homogeneous and of the same order \(n=3\). 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 &= u^{3}+u^{2}+1\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {u \left (X \right )^{3}+u \left (X \right )^{2}+1-u \left (X \right )}{X} \end {align*}
Or \[ \frac {d}{d X}u \left (X \right )-\frac {u \left (X \right )^{3}+u \left (X \right )^{2}+1-u \left (X \right )}{X} = 0 \] Or \[ -u \left (X \right )^{3}+\left (\frac {d}{d X}u \left (X \right )\right ) X -u \left (X \right )^{2}+u \left (X \right )-1 = 0 \] 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 {u^{3}+u^{2}-u +1}{X} \end {align*}
Where \(f(X)=\frac {1}{X}\) and \(g(u)=u^{3}+u^{2}-u +1\). Integrating both sides gives \begin{align*} \frac {1}{u^{3}+u^{2}-u +1} \,du &= \frac {1}{X} \,d X \\ \int { \frac {1}{u^{3}+u^{2}-u +1} \,du} &= \int {\frac {1}{X} \,d X} \\ \int _{}^{u}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a}&=\ln \left (X \right )+c_{2} \\ \end{align*} Which results in \[ \int _{}^{u}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a}=\ln \left (X \right )+c_{2} \] The solution is \[ \int _{}^{u \left (X \right )}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a} -\ln \left (X \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ \int _{}^{\frac {Y \left (X \right )}{X}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a} -\ln \left (X \right )-c_{2} = 0 \] Using the solution for \(Y(X)\) \begin {align*} \int _{}^{\frac {Y \left (X \right )}{X}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a} -\ln \left (X \right )-c_{2} = 0 \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\\ X &= x +\frac {5}{4} \end {align*}
Then the solution in \(y\) becomes \begin {align*} \int _{}^{\frac {y}{x -\frac {5}{4}}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a} -\ln \left (x -\frac {5}{4}\right )-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{\frac {y}{x -\frac {5}{4}}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a} -\ln \left (x -\frac {5}{4}\right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \int _{}^{\frac {y}{x -\frac {5}{4}}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}-\textit {\_a} +1}d \textit {\_a} -\ln \left (x -\frac {5}{4}\right )-c_{2} = 0 \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=\frac {64 x^{3}+64 x \,y^{2}+64 y^{3}-240 x^{2}-80 y^{2}+300 x -125}{64 x^{3}-240 x^{2}+300 x -125}\\ 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 (64 x^{3}+64 x \,y^{2}+64 y^{3}-240 x^{2}-80 y^{2}+300 x -125\right ) \left (b_{3}-a_{2}\right )}{64 x^{3}-240 x^{2}+300 x -125}-\frac {\left (64 x^{3}+64 x \,y^{2}+64 y^{3}-240 x^{2}-80 y^{2}+300 x -125\right )^{2} a_{3}}{\left (64 x^{3}-240 x^{2}+300 x -125\right )^{2}}-\left (\frac {192 x^{2}+64 y^{2}-480 x +300}{64 x^{3}-240 x^{2}+300 x -125}-\frac {\left (64 x^{3}+64 x \,y^{2}+64 y^{3}-240 x^{2}-80 y^{2}+300 x -125\right ) \left (192 x^{2}-480 x +300\right )}{\left (64 x^{3}-240 x^{2}+300 x -125\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (128 x y +192 y^{2}-160 y \right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{64 x^{3}-240 x^{2}+300 x -125} = 0 \end{equation} Putting the above in normal form gives \[ -\frac {4096 x^{6} a_{2}+4096 x^{6} a_{3}-4096 x^{6} b_{2}-4096 x^{6} b_{3}+8192 x^{5} y b_{2}-4096 x^{4} y^{2} a_{2}+8192 x^{4} y^{2} a_{3}+12288 x^{4} y^{2} b_{2}+4096 x^{4} y^{2} b_{3}-8192 x^{3} y^{3} a_{2}+8192 x^{3} y^{3} b_{3}-8192 x^{2} y^{4} a_{3}+8192 x \,y^{5} a_{3}+4096 y^{6} a_{3}-30720 x^{5} a_{2}-30720 x^{5} a_{3}+30720 x^{5} b_{2}+30720 x^{5} b_{3}+8192 x^{4} y b_{1}-40960 x^{4} y b_{2}-8192 x^{3} y^{2} a_{1}+10240 x^{3} y^{2} a_{2}-40960 x^{3} y^{2} a_{3}+12288 x^{3} y^{2} b_{1}-46080 x^{3} y^{2} b_{2}-20480 x^{3} y^{2} b_{3}-12288 x^{2} y^{3} a_{1}+15360 x^{2} y^{3} a_{2}-30720 x^{2} y^{3} b_{3}+20480 x \,y^{4} a_{3}-10240 y^{5} a_{3}+96000 x^{4} a_{2}+96000 x^{4} a_{3}-96000 x^{4} b_{2}-96000 x^{4} b_{3}-40960 x^{3} y b_{1}+76800 x^{3} y b_{2}+30720 x^{2} y^{2} a_{1}+76800 x^{2} y^{2} a_{3}-46080 x^{2} y^{2} b_{1}+57600 x^{2} y^{2} b_{2}+38400 x^{2} y^{2} b_{3}+30720 x \,y^{3} a_{1}+38400 x \,y^{3} b_{3}-12800 y^{4} a_{3}-160000 x^{3} a_{2}-160000 x^{3} a_{3}+160000 x^{3} b_{2}+160000 x^{3} b_{3}+76800 x^{2} y b_{1}-64000 x^{2} y b_{2}-38400 x \,y^{2} a_{1}-16000 x \,y^{2} a_{2}-64000 x \,y^{2} a_{3}+57600 x \,y^{2} b_{1}-24000 x \,y^{2} b_{2}-32000 x \,y^{2} b_{3}-19200 y^{3} a_{1}-8000 y^{3} a_{2}-16000 y^{3} b_{3}+150000 x^{2} a_{2}+150000 x^{2} a_{3}-150000 x^{2} b_{2}-150000 x^{2} b_{3}-64000 x y b_{1}+20000 x y b_{2}+16000 y^{2} a_{1}+10000 y^{2} a_{2}+20000 y^{2} a_{3}-24000 y^{2} b_{1}+10000 y^{2} b_{3}-75000 x a_{2}-75000 x a_{3}+75000 x b_{2}+75000 x b_{3}+20000 y b_{1}+15625 a_{2}+15625 a_{3}-15625 b_{2}-15625 b_{3}}{\left (64 x^{3}-240 x^{2}+300 x -125\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -4096 x^{6} a_{2}-4096 x^{6} a_{3}+4096 x^{6} b_{2}+4096 x^{6} b_{3}-8192 x^{5} y b_{2}+4096 x^{4} y^{2} a_{2}-8192 x^{4} y^{2} a_{3}-12288 x^{4} y^{2} b_{2}-4096 x^{4} y^{2} b_{3}+8192 x^{3} y^{3} a_{2}-8192 x^{3} y^{3} b_{3}+8192 x^{2} y^{4} a_{3}-8192 x \,y^{5} a_{3}-4096 y^{6} a_{3}+30720 x^{5} a_{2}+30720 x^{5} a_{3}-30720 x^{5} b_{2}-30720 x^{5} b_{3}-8192 x^{4} y b_{1}+40960 x^{4} y b_{2}+8192 x^{3} y^{2} a_{1}-10240 x^{3} y^{2} a_{2}+40960 x^{3} y^{2} a_{3}-12288 x^{3} y^{2} b_{1}+46080 x^{3} y^{2} b_{2}+20480 x^{3} y^{2} b_{3}+12288 x^{2} y^{3} a_{1}-15360 x^{2} y^{3} a_{2}+30720 x^{2} y^{3} b_{3}-20480 x \,y^{4} a_{3}+10240 y^{5} a_{3}-96000 x^{4} a_{2}-96000 x^{4} a_{3}+96000 x^{4} b_{2}+96000 x^{4} b_{3}+40960 x^{3} y b_{1}-76800 x^{3} y b_{2}-30720 x^{2} y^{2} a_{1}-76800 x^{2} y^{2} a_{3}+46080 x^{2} y^{2} b_{1}-57600 x^{2} y^{2} b_{2}-38400 x^{2} y^{2} b_{3}-30720 x \,y^{3} a_{1}-38400 x \,y^{3} b_{3}+12800 y^{4} a_{3}+160000 x^{3} a_{2}+160000 x^{3} a_{3}-160000 x^{3} b_{2}-160000 x^{3} b_{3}-76800 x^{2} y b_{1}+64000 x^{2} y b_{2}+38400 x \,y^{2} a_{1}+16000 x \,y^{2} a_{2}+64000 x \,y^{2} a_{3}-57600 x \,y^{2} b_{1}+24000 x \,y^{2} b_{2}+32000 x \,y^{2} b_{3}+19200 y^{3} a_{1}+8000 y^{3} a_{2}+16000 y^{3} b_{3}-150000 x^{2} a_{2}-150000 x^{2} a_{3}+150000 x^{2} b_{2}+150000 x^{2} b_{3}+64000 x y b_{1}-20000 x y b_{2}-16000 y^{2} a_{1}-10000 y^{2} a_{2}-20000 y^{2} a_{3}+24000 y^{2} b_{1}-10000 y^{2} b_{3}+75000 x a_{2}+75000 x a_{3}-75000 x b_{2}-75000 x b_{3}-20000 y b_{1}-15625 a_{2}-15625 a_{3}+15625 b_{2}+15625 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} -4096 a_{2} v_{1}^{6}+4096 a_{2} v_{1}^{4} v_{2}^{2}+8192 a_{2} v_{1}^{3} v_{2}^{3}-4096 a_{3} v_{1}^{6}-8192 a_{3} v_{1}^{4} v_{2}^{2}+8192 a_{3} v_{1}^{2} v_{2}^{4}-8192 a_{3} v_{1} v_{2}^{5}-4096 a_{3} v_{2}^{6}+4096 b_{2} v_{1}^{6}-8192 b_{2} v_{1}^{5} v_{2}-12288 b_{2} v_{1}^{4} v_{2}^{2}+4096 b_{3} v_{1}^{6}-4096 b_{3} v_{1}^{4} v_{2}^{2}-8192 b_{3} v_{1}^{3} v_{2}^{3}+8192 a_{1} v_{1}^{3} v_{2}^{2}+12288 a_{1} v_{1}^{2} v_{2}^{3}+30720 a_{2} v_{1}^{5}-10240 a_{2} v_{1}^{3} v_{2}^{2}-15360 a_{2} v_{1}^{2} v_{2}^{3}+30720 a_{3} v_{1}^{5}+40960 a_{3} v_{1}^{3} v_{2}^{2}-20480 a_{3} v_{1} v_{2}^{4}+10240 a_{3} v_{2}^{5}-8192 b_{1} v_{1}^{4} v_{2}-12288 b_{1} v_{1}^{3} v_{2}^{2}-30720 b_{2} v_{1}^{5}+40960 b_{2} v_{1}^{4} v_{2}+46080 b_{2} v_{1}^{3} v_{2}^{2}-30720 b_{3} v_{1}^{5}+20480 b_{3} v_{1}^{3} v_{2}^{2}+30720 b_{3} v_{1}^{2} v_{2}^{3}-30720 a_{1} v_{1}^{2} v_{2}^{2}-30720 a_{1} v_{1} v_{2}^{3}-96000 a_{2} v_{1}^{4}-96000 a_{3} v_{1}^{4}-76800 a_{3} v_{1}^{2} v_{2}^{2}+12800 a_{3} v_{2}^{4}+40960 b_{1} v_{1}^{3} v_{2}+46080 b_{1} v_{1}^{2} v_{2}^{2}+96000 b_{2} v_{1}^{4}-76800 b_{2} v_{1}^{3} v_{2}-57600 b_{2} v_{1}^{2} v_{2}^{2}+96000 b_{3} v_{1}^{4}-38400 b_{3} v_{1}^{2} v_{2}^{2}-38400 b_{3} v_{1} v_{2}^{3}+38400 a_{1} v_{1} v_{2}^{2}+19200 a_{1} v_{2}^{3}+160000 a_{2} v_{1}^{3}+16000 a_{2} v_{1} v_{2}^{2}+8000 a_{2} v_{2}^{3}+160000 a_{3} v_{1}^{3}+64000 a_{3} v_{1} v_{2}^{2}-76800 b_{1} v_{1}^{2} v_{2}-57600 b_{1} v_{1} v_{2}^{2}-160000 b_{2} v_{1}^{3}+64000 b_{2} v_{1}^{2} v_{2}+24000 b_{2} v_{1} v_{2}^{2}-160000 b_{3} v_{1}^{3}+32000 b_{3} v_{1} v_{2}^{2}+16000 b_{3} v_{2}^{3}-16000 a_{1} v_{2}^{2}-150000 a_{2} v_{1}^{2}-10000 a_{2} v_{2}^{2}-150000 a_{3} v_{1}^{2}-20000 a_{3} v_{2}^{2}+64000 b_{1} v_{1} v_{2}+24000 b_{1} v_{2}^{2}+150000 b_{2} v_{1}^{2}-20000 b_{2} v_{1} v_{2}+150000 b_{3} v_{1}^{2}-10000 b_{3} v_{2}^{2}+75000 a_{2} v_{1}+75000 a_{3} v_{1}-20000 b_{1} v_{2}-75000 b_{2} v_{1}-75000 b_{3} v_{1}-15625 a_{2}-15625 a_{3}+15625 b_{2}+15625 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} -4096 a_{3} v_{2}^{6}+10240 a_{3} v_{2}^{5}+12800 a_{3} v_{2}^{4}-20000 b_{1} v_{2}+8192 a_{3} v_{1}^{2} v_{2}^{4}-8192 a_{3} v_{1} v_{2}^{5}-8192 b_{2} v_{1}^{5} v_{2}-20480 a_{3} v_{1} v_{2}^{4}-15625 a_{2}-15625 a_{3}+15625 b_{2}+15625 b_{3}+\left (4096 a_{2}-8192 a_{3}-12288 b_{2}-4096 b_{3}\right ) v_{1}^{4} v_{2}^{2}+\left (-8192 b_{1}+40960 b_{2}\right ) v_{1}^{4} v_{2}+\left (8192 a_{2}-8192 b_{3}\right ) v_{1}^{3} v_{2}^{3}+\left (8192 a_{1}-10240 a_{2}+40960 a_{3}-12288 b_{1}+46080 b_{2}+20480 b_{3}\right ) v_{1}^{3} v_{2}^{2}+\left (40960 b_{1}-76800 b_{2}\right ) v_{1}^{3} v_{2}+\left (12288 a_{1}-15360 a_{2}+30720 b_{3}\right ) v_{1}^{2} v_{2}^{3}+\left (-30720 a_{1}-76800 a_{3}+46080 b_{1}-57600 b_{2}-38400 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (-76800 b_{1}+64000 b_{2}\right ) v_{1}^{2} v_{2}+\left (-30720 a_{1}-38400 b_{3}\right ) v_{1} v_{2}^{3}+\left (38400 a_{1}+16000 a_{2}+64000 a_{3}-57600 b_{1}+24000 b_{2}+32000 b_{3}\right ) v_{1} v_{2}^{2}+\left (64000 b_{1}-20000 b_{2}\right ) v_{1} v_{2}+\left (-4096 a_{2}-4096 a_{3}+4096 b_{2}+4096 b_{3}\right ) v_{1}^{6}+\left (30720 a_{2}+30720 a_{3}-30720 b_{2}-30720 b_{3}\right ) v_{1}^{5}+\left (-96000 a_{2}-96000 a_{3}+96000 b_{2}+96000 b_{3}\right ) v_{1}^{4}+\left (160000 a_{2}+160000 a_{3}-160000 b_{2}-160000 b_{3}\right ) v_{1}^{3}+\left (-150000 a_{2}-150000 a_{3}+150000 b_{2}+150000 b_{3}\right ) v_{1}^{2}+\left (75000 a_{2}+75000 a_{3}-75000 b_{2}-75000 b_{3}\right ) v_{1}+\left (19200 a_{1}+8000 a_{2}+16000 b_{3}\right ) v_{2}^{3}+\left (-16000 a_{1}-10000 a_{2}-20000 a_{3}+24000 b_{1}-10000 b_{3}\right ) v_{2}^{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -20480 a_{3}&=0\\ -8192 a_{3}&=0\\ -4096 a_{3}&=0\\ 8192 a_{3}&=0\\ 10240 a_{3}&=0\\ 12800 a_{3}&=0\\ -20000 b_{1}&=0\\ -8192 b_{2}&=0\\ -30720 a_{1}-38400 b_{3}&=0\\ 8192 a_{2}-8192 b_{3}&=0\\ -76800 b_{1}+64000 b_{2}&=0\\ -8192 b_{1}+40960 b_{2}&=0\\ 40960 b_{1}-76800 b_{2}&=0\\ 64000 b_{1}-20000 b_{2}&=0\\ 12288 a_{1}-15360 a_{2}+30720 b_{3}&=0\\ 19200 a_{1}+8000 a_{2}+16000 b_{3}&=0\\ -150000 a_{2}-150000 a_{3}+150000 b_{2}+150000 b_{3}&=0\\ -96000 a_{2}-96000 a_{3}+96000 b_{2}+96000 b_{3}&=0\\ -15625 a_{2}-15625 a_{3}+15625 b_{2}+15625 b_{3}&=0\\ -4096 a_{2}-4096 a_{3}+4096 b_{2}+4096 b_{3}&=0\\ 4096 a_{2}-8192 a_{3}-12288 b_{2}-4096 b_{3}&=0\\ 30720 a_{2}+30720 a_{3}-30720 b_{2}-30720 b_{3}&=0\\ 75000 a_{2}+75000 a_{3}-75000 b_{2}-75000 b_{3}&=0\\ 160000 a_{2}+160000 a_{3}-160000 b_{2}-160000 b_{3}&=0\\ -30720 a_{1}-76800 a_{3}+46080 b_{1}-57600 b_{2}-38400 b_{3}&=0\\ -16000 a_{1}-10000 a_{2}-20000 a_{3}+24000 b_{1}-10000 b_{3}&=0\\ 8192 a_{1}-10240 a_{2}+40960 a_{3}-12288 b_{1}+46080 b_{2}+20480 b_{3}&=0\\ 38400 a_{1}+16000 a_{2}+64000 a_{3}-57600 b_{1}+24000 b_{2}+32000 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=-\frac {5 b_{3}}{4}\\ 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 &= x -\frac {5}{4} \\ \eta &= y \\ \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 {y}{x -\frac {5}{4}}\\ &= \frac {4 y}{4 x -5} \end {align*}
This is easily solved to give \begin {align*} y = c_{1} \left (4 x -5\right ) \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {y}{4 x -5} \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{x -\frac {5}{4}} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= \ln \left (4 x -5\right ) \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) &= \frac {64 x^{3}+64 x \,y^{2}+64 y^{3}-240 x^{2}-80 y^{2}+300 x -125}{64 x^{3}-240 x^{2}+300 x -125} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= -\frac {4 y}{\left (4 x -5\right )^{2}}\\ R_{y} &= \frac {1}{4 x -5}\\ S_{x} &= \frac {4}{4 x -5}\\ 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 {4 \left (4 x -5\right )^{3}}{64 x^{3}+\left (-64 y -240\right ) x^{2}+\left (64 y^{2}+160 y +300\right ) x +64 y^{3}-80 y^{2}-100 y -125}\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 {4}{64 R^{3}+16 R^{2}-4 R +1} \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 ) = \int \frac {4}{64 R^{3}+16 R^{2}-4 R +1}d R +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*} \ln \left (4 x -5\right ) = \int _{}^{\frac {y}{4 x -5}}\frac {4}{64 \textit {\_a}^{3}+16 \textit {\_a}^{2}-4 \textit {\_a} +1}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} \ln \left (4 x -5\right ) = \int _{}^{\frac {y}{4 x -5}}\frac {4}{64 \textit {\_a}^{3}+16 \textit {\_a}^{2}-4 \textit {\_a} +1}d \textit {\_a} +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 \((R,S)\) |
| \( \frac {dy}{dx} = \frac {64 x^{3}+64 x \,y^{2}+64 y^{3}-240 x^{2}-80 y^{2}+300 x -125}{64 x^{3}-240 x^{2}+300 x -125}\) |
|
\( \frac {d S}{d R} = \frac {4}{64 R^{3}+16 R^{2}-4 R +1}\) |
| |
\(\!\begin {aligned} R&= \frac {y}{4 x -5}\\ S&= \ln \left (4 x -5\right ) \end {aligned} \) |
|
Summary
The solution(s) found are the following \begin{align*} \tag{1} \ln \left (4 x -5\right ) &= \int _{}^{\frac {y}{4 x -5}}\frac {4}{64 \textit {\_a}^{3}+16 \textit {\_a}^{2}-4 \textit {\_a} +1}d \textit {\_a} +c_{1} \\ \end{align*}
Verification of solutions
\[ \ln \left (4 x -5\right ) = \int _{}^{\frac {y}{4 x -5}}\frac {4}{64 \textit {\_a}^{3}+16 \textit {\_a}^{2}-4 \textit {\_a} +1}d \textit {\_a} +c_{1} \] Verified OK.
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=1+\frac {64 y^{3}}{64 x^{3}-240 x^{2}+300 x -125}+\frac {\left (64 x -80\right ) y^{2}}{64 x^{3}-240 x^{2}+300 x -125}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 1\\ f_1(x) &= 0\\ f_2(x) &= \frac {64 x}{64 x^{3}-240 x^{2}+300 x -125}-\frac {80}{64 x^{3}-240 x^{2}+300 x -125}\\ f_3(x) &= \frac {64}{64 x^{3}-240 x^{2}+300 x -125} \end {align*}
Since \(f_2(x)=\frac {64 x}{64 x^{3}-240 x^{2}+300 x -125}-\frac {80}{64 x^{3}-240 x^{2}+300 x -125}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {\frac {64 x}{64 x^{3}-240 x^{2}+300 x -125}-\frac {80}{64 x^{3}-240 x^{2}+300 x -125}}{\frac {192}{64 x^{3}-240 x^{2}+300 x -125}} \right ) \\ &= u \left (x \right )-\frac {x}{3}+\frac {5}{12} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {64 u \left (x \right )^{3}}{64 x^{3}-240 x^{2}+300 x -125}-\frac {64 u \left (x \right ) x^{2}}{3 \left (64 x^{3}-240 x^{2}+300 x -125\right )}+\frac {2432 x^{3}}{27 \left (64 x^{3}-240 x^{2}+300 x -125\right )}+\frac {160 u \left (x \right ) x}{3 \left (64 x^{3}-240 x^{2}+300 x -125\right )}-\frac {3040 x^{2}}{9 \left (64 x^{3}-240 x^{2}+300 x -125\right )}-\frac {100 u \left (x \right )}{3 \left (64 x^{3}-240 x^{2}+300 x -125\right )}+\frac {3800 x}{9 \left (64 x^{3}-240 x^{2}+300 x -125\right )}-\frac {4750}{27 \left (64 x^{3}-240 x^{2}+300 x -125\right )}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Let \(Y = u +y_{0}\) 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 {64 \left (Y \left (X \right )+y_{0} \right )^{3}-\frac {64 \left (Y \left (X \right )+y_{0} \right ) \left (x_{0} +X \right )^{2}}{3}+\frac {2432 \left (x_{0} +X \right )^{3}}{27}+\frac {160 \left (Y \left (X \right )+y_{0} \right ) \left (x_{0} +X \right )}{3}-\frac {3040 \left (x_{0} +X \right )^{2}}{9}-\frac {100 Y \left (X \right )}{3}-\frac {100 y_{0}}{3}+\frac {3800 x_{0}}{9}+\frac {3800 X}{9}-\frac {4750}{27}}{64 \left (x_{0} +X \right )^{3}-240 \left (x_{0} +X \right )^{2}+300 x_{0} +300 X -125} \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 {5}{4}}\\ y_{0}&=0 \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 {1216 X^{3}-288 Y \left (X \right ) X^{2}+864 Y \left (X \right )^{3}}{864 X^{3}} \end {align*}
In canonical form, the ODE is \begin {align*} Y' &= F(X,Y)\\ &= \frac {38 X^{3}-9 Y \,X^{2}+27 Y^{3}}{27 X^{3}}\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 \(M=38 X^{3}-9 Y \,X^{2}+27 Y^{3}\) and \(N=27 X^{3}\) are both homogeneous and of the same order \(n=3\). 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 {38}{27}-\frac {1}{3} u +u^{3}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {38}{27}-\frac {4 u \left (X \right )}{3}+u \left (X \right )^{3}}{X} \end {align*}
Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {38}{27}-\frac {4 u \left (X \right )}{3}+u \left (X \right )^{3}}{X} = 0 \] Or \[ -27 u \left (X \right )^{3}+27 \left (\frac {d}{d X}u \left (X \right )\right ) X +36 u \left (X \right )-38 = 0 \] 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 {u^{3}-\frac {4}{3} u +\frac {38}{27}}{X} \end {align*}
Where \(f(X)=\frac {1}{X}\) and \(g(u)=u^{3}-\frac {4}{3} u +\frac {38}{27}\). Integrating both sides gives \begin{align*} \frac {1}{u^{3}-\frac {4}{3} u +\frac {38}{27}} \,du &= \frac {1}{X} \,d X \\ \int { \frac {1}{u^{3}-\frac {4}{3} u +\frac {38}{27}} \,du} &= \int {\frac {1}{X} \,d X} \\ \int _{}^{u}\frac {1}{\textit {\_a}^{3}-\frac {4}{3} \textit {\_a} +\frac {38}{27}}d \textit {\_a}&=\ln \left (X \right )+c_{4} \\ \end{align*} Which results in \[ \int _{}^{u}\frac {1}{\textit {\_a}^{3}-\frac {4}{3} \textit {\_a} +\frac {38}{27}}d \textit {\_a}=\ln \left (X \right )+c_{4} \] The solution is \[ \int _{}^{u \left (X \right )}\frac {1}{\textit {\_a}^{3}-\frac {4}{3} \textit {\_a} +\frac {38}{27}}d \textit {\_a} -\ln \left (X \right )-c_{4} = 0 \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ \int _{}^{\frac {Y \left (X \right )}{X}}\frac {1}{\textit {\_a}^{3}-\frac {4}{3} \textit {\_a} +\frac {38}{27}}d \textit {\_a} -\ln \left (X \right )-c_{4} = 0 \] Using the solution for \(Y(X)\) \begin {align*} 27 \left (\int _{}^{\frac {Y \left (X \right )}{X}}\frac {1}{27 \textit {\_a}^{3}-36 \textit {\_a} +38}d \textit {\_a} \right )-\ln \left (X \right )-c_{4} = 0 \end {align*}
And replacing back terms in the above solution using \begin {align*} Y &= u +y_{0}\\ X &= x +x_{0} \end {align*}
Or \begin {align*} Y &= u\\ X &= x +\frac {5}{4} \end {align*}
Then the solution in \(u \left (x \right )\) becomes \begin {align*} 27 \left (\int _{}^{\frac {u \left (x \right )}{x -\frac {5}{4}}}\frac {1}{27 \textit {\_a}^{3}-36 \textit {\_a} +38}d \textit {\_a} \right )-\ln \left (x -\frac {5}{4}\right )-c_{4} = 0 \end {align*}
Substituting \(u=y-\frac {\left (\frac {64 x}{64 x^{3}-240 x^{2}+300 x -125}-\frac {80}{64 x^{3}-240 x^{2}+300 x -125}\right ) \left (64 x^{3}-240 x^{2}+300 x -125\right )}{192}\) in the above solution gives \begin {align*} 27 \left (\int _{}^{\frac {y-\frac {\left (\frac {64 x}{64 x^{3}-240 x^{2}+300 x -125}-\frac {80}{64 x^{3}-240 x^{2}+300 x -125}\right ) \left (64 x^{3}-240 x^{2}+300 x -125\right )}{192}}{x -\frac {5}{4}}}\frac {1}{27 \textit {\_a}^{3}-36 \textit {\_a} +38}d \textit {\_a} \right )-\ln \left (x -\frac {5}{4}\right )-c_{4} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} 27 \left (\int _{}^{\frac {y-\frac {\left (\frac {64 x}{64 x^{3}-240 x^{2}+300 x -125}-\frac {80}{64 x^{3}-240 x^{2}+300 x -125}\right ) \left (64 x^{3}-240 x^{2}+300 x -125\right )}{192}}{x -\frac {5}{4}}}\frac {1}{27 \textit {\_a}^{3}-36 \textit {\_a} +38}d \textit {\_a} \right )-\ln \left (x -\frac {5}{4}\right )-c_{4} &= 0 \\ \end{align*}
Verification of solutions
\[ 27 \left (\int _{}^{\frac {y-\frac {\left (\frac {64 x}{64 x^{3}-240 x^{2}+300 x -125}-\frac {80}{64 x^{3}-240 x^{2}+300 x -125}\right ) \left (64 x^{3}-240 x^{2}+300 x -125\right )}{192}}{x -\frac {5}{4}}}\frac {1}{27 \textit {\_a}^{3}-36 \textit {\_a} +38}d \textit {\_a} \right )-\ln \left (x -\frac {5}{4}\right )-c_{4} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -64 y^{\prime } x^{3}+64 y^{3}+64 y^{2} x +240 y^{\prime } x^{2}+64 x^{3}-80 y^{2}-300 y^{\prime } x -240 x^{2}+125 y^{\prime }+300 x -125=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {125-300 x +240 x^{2}-64 x^{3}+80 y^{2}-64 y^{2} x -64 y^{3}}{-64 x^{3}+240 x^{2}-300 x +125} \end {array} \]
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: 0.015 (sec). Leaf size: 41
dsolve(diff(y(x),x) = (-125+300*x-240*x^2+64*x^3-80*y(x)^2+64*x*y(x)^2+64*y(x)^3)/(4*x-5)^3,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a}^{3}-\textit {\_a}^{2}-\textit {\_a} -1}d \textit {\_a} \right )+\ln \left (4 x -5\right )+c_{1} \right ) \left (4 x -5\right )}{4} \]
✓ Solution by Mathematica
Time used: 0.247 (sec). Leaf size: 128
DSolve[y'[x] == (-125 + 300*x - 240*x^2 + 64*x^3 - 80*y[x]^2 + 64*x*y[x]^2 + 64*y[x]^3)/(-5 + 4*x)^3,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {19}{3} \text {RootSum}\left [-19 \text {$\#$1}^3+6 \sqrt [3]{38} \text {$\#$1}-19\&,\frac {\log \left (\frac {\frac {192 y(x)}{(4 x-5)^3}+\frac {16}{(4 x-5)^2}}{16 \sqrt [3]{38} \sqrt [3]{\frac {1}{(4 x-5)^6}}}-\text {$\#$1}\right )}{2 \sqrt [3]{38}-19 \text {$\#$1}^2}\&\right ]=\frac {1}{9} 38^{2/3} \left (\frac {1}{(5-4 x)^6}\right )^{2/3} (5-4 x)^4 \log (5-4 x)+c_1,y(x)\right ] \]