Internal problem ID [9124]
Internal file name [OUTPUT/8059_Monday_June_06_2022_01_31_40_AM_43688869/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 790.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {2 x \ln \left (\frac {1}{x -1}\right )-\coth \left (\frac {x +1}{x -1}\right )+\coth \left (\frac {x +1}{x -1}\right ) y^{2}-2 \coth \left (\frac {x +1}{x -1}\right ) x^{2} y+\coth \left (\frac {x +1}{x -1}\right ) x^{4}}{\ln \left (\frac {1}{x -1}\right )}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {2 x \ln \left (\frac {1}{x -1}\right )-\coth \left (\frac {x +1}{x -1}\right )+\coth \left (\frac {x +1}{x -1}\right ) y^{2}-2 \coth \left (\frac {x +1}{x -1}\right ) x^{2} y +\coth \left (\frac {x +1}{x -1}\right ) x^{4}}{\ln \left (\frac {1}{x -1}\right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = 2 x -\frac {\cosh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )}-\frac {\sinh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )}+\frac {y^{2} \cosh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )}+\frac {y^{2} \sinh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )}-\frac {2 x^{2} y \cosh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )}-\frac {2 x^{2} y \sinh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )}+\frac {x^{4} \cosh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )}+\frac {x^{4} \sinh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right ) \left (\sinh \left (\frac {x}{x -1}\right ) \cosh \left (\frac {1}{x -1}\right )+\cosh \left (\frac {x}{x -1}\right ) \sinh \left (\frac {1}{x -1}\right )\right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\coth \left (\frac {x +1}{x -1}\right ) x^{4}+2 x \ln \left (\frac {1}{x -1}\right )-\coth \left (\frac {x +1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right )}\), \(f_1(x)=-\frac {2 \coth \left (\frac {x +1}{x -1}\right ) x^{2}}{\ln \left (\frac {1}{x -1}\right )}\) and \(f_2(x)=\frac {\coth \left (\frac {x +1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\coth \left (\frac {x +1}{x -1}\right ) u}{\ln \left (\frac {1}{x -1}\right )}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=\frac {\coth \left (\frac {x +1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right )^{2} \left (x -1\right )}+\frac {\left (\frac {1}{x -1}-\frac {x +1}{\left (x -1\right )^{2}}\right ) \left (1-\coth \left (\frac {x +1}{x -1}\right )^{2}\right )}{\ln \left (\frac {1}{x -1}\right )}\\ f_1 f_2 &=-\frac {2 \coth \left (\frac {x +1}{x -1}\right )^{2} x^{2}}{\ln \left (\frac {1}{x -1}\right )^{2}}\\ f_2^2 f_0 &=\frac {\coth \left (\frac {x +1}{x -1}\right )^{2} \left (\coth \left (\frac {x +1}{x -1}\right ) x^{4}+2 x \ln \left (\frac {1}{x -1}\right )-\coth \left (\frac {x +1}{x -1}\right )\right )}{\ln \left (\frac {1}{x -1}\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\coth \left (\frac {x +1}{x -1}\right ) u^{\prime \prime }\left (x \right )}{\ln \left (\frac {1}{x -1}\right )}-\left (\frac {\coth \left (\frac {x +1}{x -1}\right )}{\ln \left (\frac {1}{x -1}\right )^{2} \left (x -1\right )}+\frac {\left (\frac {1}{x -1}-\frac {x +1}{\left (x -1\right )^{2}}\right ) \left (1-\coth \left (\frac {x +1}{x -1}\right )^{2}\right )}{\ln \left (\frac {1}{x -1}\right )}-\frac {2 \coth \left (\frac {x +1}{x -1}\right )^{2} x^{2}}{\ln \left (\frac {1}{x -1}\right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\coth \left (\frac {x +1}{x -1}\right )^{2} \left (\coth \left (\frac {x +1}{x -1}\right ) x^{4}+2 x \ln \left (\frac {1}{x -1}\right )-\coth \left (\frac {x +1}{x -1}\right )\right ) u \left (x \right )}{\ln \left (\frac {1}{x -1}\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives Unable to solve. Terminating.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -\coth \left (\frac {x +1}{x -1}\right ) x^{4}+2 \coth \left (\frac {x +1}{x -1}\right ) x^{2} y-\coth \left (\frac {x +1}{x -1}\right ) y^{2}+y^{\prime } \ln \left (\frac {1}{x -1}\right )-2 x \ln \left (\frac {1}{x -1}\right )+\coth \left (\frac {x +1}{x -1}\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 x \ln \left (\frac {1}{x -1}\right )-\coth \left (\frac {x +1}{x -1}\right )+\coth \left (\frac {x +1}{x -1}\right ) y^{2}-2 \coth \left (\frac {x +1}{x -1}\right ) x^{2} y+\coth \left (\frac {x +1}{x -1}\right ) x^{4}}{\ln \left (\frac {1}{x -1}\right )} \end {array} \]
✗ Solution by Maple
dsolve(diff(y(x),x) = (2*x*ln(1/(x-1))-coth((x+1)/(x-1))+coth((x+1)/(x-1))*y(x)^2-2*coth((x+1)/(x-1))*x^2*y(x)+coth((x+1)/(x-1))*x^4)/ln(1/(x-1)),y(x), singsol=all)
\[ \text {No solution found} \]
✓ Solution by Mathematica
Time used: 106.677 (sec). Leaf size: 228
DSolve[y'[x] == (-Coth[(1 + x)/(-1 + x)] + x^4*Coth[(1 + x)/(-1 + x)] + 2*x*Log[(-1 + x)^(-1)] - 2*x^2*Coth[(1 + x)/(-1 + x)]*y[x] + Coth[(1 + x)/(-1 + x)]*y[x]^2)/Log[(-1 + x)^(-1)],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\exp \left (\int _1^x\frac {2 \coth \left (\frac {K[5]+1}{K[5]-1}\right )}{\log \left (\frac {1}{K[5]-1}\right )}dK[5]\right )}{-\int _1^x\frac {\exp \left (\int _1^{K[6]}\frac {2 \coth \left (\frac {K[5]+1}{K[5]-1}\right )}{\log \left (\frac {1}{K[5]-1}\right )}dK[5]\right ) \coth \left (\frac {K[6]+1}{K[6]-1}\right )}{\log \left (\frac {1}{K[6]-1}\right )}dK[6]+c_1}+x^2+1 \\ y(x)\to x^2+1 \\ y(x)\to -\frac {\exp \left (\int _1^x\frac {2 \coth \left (\frac {K[5]+1}{K[5]-1}\right )}{\log \left (\frac {1}{K[5]-1}\right )}dK[5]\right )}{\int _1^x\frac {\exp \left (\int _1^{K[6]}\frac {2 \coth \left (\frac {K[5]+1}{K[5]-1}\right )}{\log \left (\frac {1}{K[5]-1}\right )}dK[5]\right ) \coth \left (\frac {K[6]+1}{K[6]-1}\right )}{\log \left (\frac {1}{K[6]-1}\right )}dK[6]}+x^2+1 \\ \end{align*}