Internal problem ID [9318]
Internal file name [OUTPUT/8254_Monday_June_06_2022_02_32_32_AM_26085347/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 985.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[_rational, [_1st_order, `_with_symmetry_[F(x),G(x)]`], _Abel]
\[ \boxed {y^{\prime }-\frac {\left (y x +1\right ) \left (y^{2} x^{2}+y x^{2}+2 y x +x^{2}+x +1\right )}{x^{5}}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {x^{3} y^{3}+y^{2} x^{3}+y \,x^{3}+3 y^{2} x^{2}+2 x^{2} y +x^{2}+3 y x +x +1}{x^{5}}\\ 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 2 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} 2 x b_{4}+y b_{5}+b_{2}+\frac {\left (x^{3} y^{3}+y^{2} x^{3}+y \,x^{3}+3 y^{2} x^{2}+2 x^{2} y +x^{2}+3 y x +x +1\right ) \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{x^{5}}-\frac {\left (x^{3} y^{3}+y^{2} x^{3}+y \,x^{3}+3 y^{2} x^{2}+2 x^{2} y +x^{2}+3 y x +x +1\right )^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{x^{10}}-\left (\frac {3 y^{3} x^{2}+3 y^{2} x^{2}+3 x^{2} y +6 y^{2} x +4 y x +2 x +3 y +1}{x^{5}}-\frac {5 \left (x^{3} y^{3}+y^{2} x^{3}+y \,x^{3}+3 y^{2} x^{2}+2 x^{2} y +x^{2}+3 y x +x +1\right )}{x^{6}}\right ) \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\frac {\left (3 y^{2} x^{3}+2 y \,x^{3}+x^{3}+6 x^{2} y +2 x^{2}+3 x \right ) \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right )}{x^{5}} = 0 \end{equation} Putting the above in normal form gives \[ \text {Expression too large to display} \] Setting the numerator to zero gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (-6 a_{3}-4 a_{6}\right ) v_{1} v_{2}+\left (-2 b_{4}+b_{5}\right ) v_{1}^{10} v_{2}+\left (-3 b_{2}-b_{5}\right ) v_{1}^{9} v_{2}^{2}+\left (-2 b_{2}-6 b_{4}\right ) v_{1}^{9} v_{2}+\left (a_{5}-b_{6}\right ) v_{1}^{8} v_{2}^{4}+\left (a_{2}+a_{5}-2 b_{3}\right ) v_{1}^{8} v_{2}^{3}+\left (a_{2}+3 a_{4}+a_{5}-3 b_{1}-b_{3}-3 b_{5}+b_{6}\right ) v_{1}^{8} v_{2}^{2}+\left (a_{2}+2 a_{4}-2 b_{1}-6 b_{2}\right ) v_{1}^{8} v_{2}+\left (-2 a_{5}+2 a_{6}\right ) v_{1}^{7} v_{2}^{5}+\left (2 a_{3}-3 a_{5}+2 a_{6}\right ) v_{1}^{7} v_{2}^{4}+\left (2 a_{1}+2 a_{3}+4 a_{5}+2 a_{6}\right ) v_{1}^{7} v_{2}^{3}+\left (2 a_{1}+6 a_{2}+2 a_{3}+3 a_{5}-3 b_{3}+2 b_{6}\right ) v_{1}^{7} v_{2}^{2}+\left (2 a_{1}+4 a_{2}+6 a_{4}+2 a_{5}-6 b_{1}+2 b_{6}\right ) v_{1}^{7} v_{2}+\left (-a_{3}-4 a_{6}\right ) v_{1}^{6} v_{2}^{6}+\left (-2 a_{3}-6 a_{5}-6 a_{6}\right ) v_{1}^{6} v_{2}^{5}+\left (-3 a_{3}-10 a_{5}+5 a_{6}\right ) v_{1}^{6} v_{2}^{4}+\left (7 a_{3}-12 a_{5}+4 a_{6}\right ) v_{1}^{6} v_{2}^{3}+\left (9 a_{1}+5 a_{3}+3 a_{5}+3 a_{6}+3 b_{6}\right ) v_{1}^{6} v_{2}^{2}+\left (6 a_{1}+9 a_{2}+3 a_{3}+a_{5}+2 b_{6}\right ) v_{1}^{6} v_{2}+\left (-6 a_{3}-20 a_{6}\right ) v_{1}^{5} v_{2}^{5}+\left (-10 a_{3}-15 a_{5}-24 a_{6}\right ) v_{1}^{5} v_{2}^{4}+\left (-12 a_{3}-20 a_{5}\right ) v_{1}^{5} v_{2}^{3}+\left (6 a_{3}-18 a_{5}\right ) v_{1}^{5} v_{2}^{2}+\left (12 a_{1}+2 a_{3}-2 a_{5}+2 b_{6}\right ) v_{1}^{5} v_{2}+\left (-15 a_{3}-40 a_{6}\right ) v_{1}^{4} v_{2}^{4}+\left (-20 a_{3}-20 a_{5}-36 a_{6}\right ) v_{1}^{4} v_{2}^{3}+\left (-18 a_{3}-20 a_{5}-7 a_{6}\right ) v_{1}^{4} v_{2}^{2}+\left (-a_{3}-12 a_{5}-2 a_{6}\right ) v_{1}^{4} v_{2}+\left (-20 a_{3}-40 a_{6}\right ) v_{1}^{3} v_{2}^{3}+\left (-20 a_{3}-15 a_{5}-24 a_{6}\right ) v_{1}^{3} v_{2}^{2}+\left (-12 a_{3}-10 a_{5}-4 a_{6}\right ) v_{1}^{3} v_{2}+\left (-15 a_{3}-20 a_{6}\right ) v_{1}^{2} v_{2}^{2}+\left (-10 a_{3}-6 a_{5}-6 a_{6}\right ) v_{1}^{2} v_{2}-a_{3}+\left (b_{2}-b_{4}\right ) v_{1}^{10}-30 a_{6} v_{1}^{2} v_{2}^{3}-12 a_{6} v_{1} v_{2}^{2}-40 a_{6} v_{1}^{3} v_{2}^{4}-30 a_{6} v_{1}^{4} v_{2}^{5}-2 a_{6} v_{1}^{6} v_{2}^{7}-3 b_{4} v_{1}^{10} v_{2}^{2}-2 b_{5} v_{1}^{9} v_{2}^{3}-12 a_{6} v_{1}^{5} v_{2}^{6}+2 b_{4} v_{1}^{11}-2 a_{6} v_{2}-a_{5} v_{1}^{7} v_{2}^{6}+\left (3 a_{1}+3 a_{2}+3 a_{4}-3 b_{1}+b_{3}+b_{5}\right ) v_{1}^{6}+\left (4 a_{1}+4 a_{2}-a_{5}+b_{3}\right ) v_{1}^{5}+\left (-b_{2}-2 b_{4}\right ) v_{1}^{9}+\left (a_{4}-b_{1}-2 b_{2}-3 b_{4}+b_{5}\right ) v_{1}^{8}+\left (2 a_{2}+2 a_{4}-2 b_{1}-3 b_{2}+b_{3}+b_{5}\right ) v_{1}^{7}+\left (5 a_{1}-a_{3}-2 a_{5}\right ) v_{1}^{4}+\left (-2 a_{3}-3 a_{5}\right ) v_{1}^{3}+\left (-3 a_{3}-2 a_{5}\right ) v_{1}^{2}+\left (-2 a_{3}-a_{5}\right ) v_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -a_{3}&=0\\ -a_{5}&=0\\ -40 a_{6}&=0\\ -30 a_{6}&=0\\ -12 a_{6}&=0\\ -2 a_{6}&=0\\ -3 b_{4}&=0\\ 2 b_{4}&=0\\ -2 b_{5}&=0\\ -20 a_{3}-40 a_{6}&=0\\ -15 a_{3}-40 a_{6}&=0\\ -15 a_{3}-20 a_{6}&=0\\ -12 a_{3}-20 a_{5}&=0\\ -6 a_{3}-20 a_{6}&=0\\ -6 a_{3}-4 a_{6}&=0\\ -3 a_{3}-2 a_{5}&=0\\ -2 a_{3}-3 a_{5}&=0\\ -2 a_{3}-a_{5}&=0\\ -a_{3}-4 a_{6}&=0\\ 6 a_{3}-18 a_{5}&=0\\ -2 a_{5}+2 a_{6}&=0\\ a_{5}-b_{6}&=0\\ -3 b_{2}-b_{5}&=0\\ -2 b_{2}-6 b_{4}&=0\\ -b_{2}-2 b_{4}&=0\\ b_{2}-b_{4}&=0\\ -2 b_{4}+b_{5}&=0\\ 5 a_{1}-a_{3}-2 a_{5}&=0\\ a_{2}+a_{5}-2 b_{3}&=0\\ -20 a_{3}-20 a_{5}-36 a_{6}&=0\\ -20 a_{3}-15 a_{5}-24 a_{6}&=0\\ -18 a_{3}-20 a_{5}-7 a_{6}&=0\\ -12 a_{3}-10 a_{5}-4 a_{6}&=0\\ -10 a_{3}-15 a_{5}-24 a_{6}&=0\\ -10 a_{3}-6 a_{5}-6 a_{6}&=0\\ -3 a_{3}-10 a_{5}+5 a_{6}&=0\\ -2 a_{3}-6 a_{5}-6 a_{6}&=0\\ -a_{3}-12 a_{5}-2 a_{6}&=0\\ 2 a_{3}-3 a_{5}+2 a_{6}&=0\\ 7 a_{3}-12 a_{5}+4 a_{6}&=0\\ 2 a_{1}+2 a_{3}+4 a_{5}+2 a_{6}&=0\\ 4 a_{1}+4 a_{2}-a_{5}+b_{3}&=0\\ 12 a_{1}+2 a_{3}-2 a_{5}+2 b_{6}&=0\\ a_{2}+2 a_{4}-2 b_{1}-6 b_{2}&=0\\ 6 a_{1}+9 a_{2}+3 a_{3}+a_{5}+2 b_{6}&=0\\ 9 a_{1}+5 a_{3}+3 a_{5}+3 a_{6}+3 b_{6}&=0\\ a_{4}-b_{1}-2 b_{2}-3 b_{4}+b_{5}&=0\\ 2 a_{1}+4 a_{2}+6 a_{4}+2 a_{5}-6 b_{1}+2 b_{6}&=0\\ 2 a_{1}+6 a_{2}+2 a_{3}+3 a_{5}-3 b_{3}+2 b_{6}&=0\\ 3 a_{1}+3 a_{2}+3 a_{4}-3 b_{1}+b_{3}+b_{5}&=0\\ 2 a_{2}+2 a_{4}-2 b_{1}-3 b_{2}+b_{3}+b_{5}&=0\\ a_{2}+3 a_{4}+a_{5}-3 b_{1}-b_{3}-3 b_{5}+b_{6}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=b_{1}\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=b_{1}\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=0\\ b_{6}&=0 \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= x^{2} \\ \eta &= 1 \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
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 }&=\frac {y^{3}}{x^{2}}+\frac {\left (x^{3}+3 x^{2}\right ) y^{2}}{x^{5}}+\frac {\left (x^{3}+2 x^{2}+3 x \right ) y}{x^{5}}+\frac {x^{2}+x +1}{x^{5}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {1}{x^{3}}+\frac {1}{x^{4}}+\frac {1}{x^{5}}\\ f_1(x) &= \frac {1}{x^{2}}+\frac {2}{x^{3}}+\frac {3}{x^{4}}\\ f_2(x) &= \frac {1}{x^{2}}+\frac {3}{x^{3}}\\ f_3(x) &= \frac {1}{x^{2}} \end {align*}
Since \(f_2(x)=\frac {1}{x^{2}}+\frac {3}{x^{3}}\) 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 {1}{x^{2}}+\frac {3}{x^{3}}}{\frac {3}{x^{2}}} \right ) \\ &= u \left (x \right )-\frac {x +3}{3 x} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{3}}{x^{2}}+\frac {2 u \left (x \right )}{3 x^{2}}-\frac {34}{27 x^{2}}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u^{3}+\frac {2}{3} u -\frac {34}{27}}{x^{2}} \end {align*}
Where \(f(x)=\frac {1}{x^{2}}\) and \(g(u)=u^{3}+\frac {2}{3} u -\frac {34}{27}\). Integrating both sides gives \begin{align*} \frac {1}{u^{3}+\frac {2}{3} u -\frac {34}{27}} \,du &= \frac {1}{x^{2}} \,d x \\ \int { \frac {1}{u^{3}+\frac {2}{3} u -\frac {34}{27}} \,du} &= \int {\frac {1}{x^{2}} \,d x} \\ \int _{}^{u}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {34}{27}}d \textit {\_a}&=-\frac {1}{x}+c_{2} \\ \end{align*} Which results in \[ \int _{}^{u}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {34}{27}}d \textit {\_a}=-\frac {1}{x}+c_{2} \] The solution is \[ \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {34}{27}}d \textit {\_a} +\frac {1}{x}-c_{2} = 0 \] Substituting \(u=y+\frac {\left (\frac {1}{x^{2}}+\frac {3}{x^{3}}\right ) x^{2}}{3}\) in the above solution gives \begin {align*} \int _{}^{y+\frac {\left (\frac {1}{x^{2}}+\frac {3}{x^{3}}\right ) x^{2}}{3}}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {34}{27}}d \textit {\_a} +\frac {1}{x}-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y+\frac {\left (\frac {1}{x^{2}}+\frac {3}{x^{3}}\right ) x^{2}}{3}}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {34}{27}}d \textit {\_a} +\frac {1}{x}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \int _{}^{y+\frac {\left (\frac {1}{x^{2}}+\frac {3}{x^{3}}\right ) x^{2}}{3}}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {34}{27}}d \textit {\_a} +\frac {1}{x}-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x^{5}-y^{3} x^{3}-y^{2} x^{3}-3 y^{2} x^{2}-y x^{3}-2 y x^{2}-3 y x -x^{2}-x -1=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3} x^{3}+y^{2} x^{3}+3 y^{2} x^{2}+y x^{3}+2 y x^{2}+3 y x +x^{2}+x +1}{x^{5}} \end {array} \]
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 Chini differential order: 1; looking for linear symmetries trying exact trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 43
dsolve(diff(y(x),x) = (x*y(x)+1)*(x^2*y(x)^2+x^2*y(x)+2*x*y(x)+1+x+x^2)/x^5,y(x), singsol=all)
\[ y \left (x \right ) = \frac {17 \operatorname {RootOf}\left (162 \left (\int _{}^{\textit {\_Z}}\frac {1}{289 \textit {\_a}^{3}+54 \textit {\_a} -54}d \textit {\_a} \right ) x +3 c_{1} x +2\right ) x -3 x -9}{9 x} \]
✓ Solution by Mathematica
Time used: 0.257 (sec). Leaf size: 103
DSolve[y'[x] == ((1 + x*y[x])*(1 + x + x^2 + 2*x*y[x] + x^2*y[x] + x^2*y[x]^2))/x^5,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {17}{3} \text {RootSum}\left [-17 \text {$\#$1}^3+3 \sqrt [3]{-34} \text {$\#$1}-17\&,\frac {\log \left (\frac {\frac {x+3}{x^3}+\frac {3 y(x)}{x^2}}{\sqrt [3]{34} \sqrt [3]{-\frac {1}{x^6}}}-\text {$\#$1}\right )}{\sqrt [3]{-34}-17 \text {$\#$1}^2}\&\right ]=-\frac {1}{9} 34^{2/3} \left (-\frac {1}{x^6}\right )^{2/3} x^3+c_1,y(x)\right ] \]