Internal problem ID [8368]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 788.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [_Bernoulli]
Solve \begin {gather*} \boxed {y^{\prime }+\frac {y \left (\ln \left (x -1\right )+\coth \left (x +1\right ) x -\coth \left (x +1\right ) x^{2} y\right )}{x \ln \left (x -1\right )}=0} \end {gather*}
✓ Solution by Maple
Time used: 0.018 (sec). Leaf size: 106
dsolve(diff(y(x),x) = -y(x)*(ln(x-1)+coth(x+1)*x-coth(x+1)*x^2*y(x))/x/ln(x-1),y(x), singsol=all)
\[ y \relax (x ) = \frac {{\mathrm e}^{\int -\frac {x \cosh \left (x +1\right )+\ln \left (x -1\right ) \sinh \left (x +1\right )}{x \ln \left (x -1\right ) \sinh \left (x +1\right )}d x}}{c_{1}+\int -\frac {{\mathrm e}^{\int -\frac {x \cosh \left (x +1\right )+\ln \left (x -1\right ) \sinh \left (x +1\right )}{x \ln \left (x -1\right ) \sinh \left (x +1\right )}d x} x \cosh \left (x +1\right )}{\ln \left (x -1\right ) \sinh \left (x +1\right )}d x} \]
✓ Solution by Mathematica
Time used: 34.729 (sec). Leaf size: 350
DSolve[y'[x] == -((y[x]*(x*Coth[1 + x] + Log[-1 + x] - x^2*Coth[1 + x]*y[x]))/(x*Log[-1 + x])),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\exp \left (\int _1^x\frac {-K[1]+\log (K[1]-1)-e^{2 K[1]+2} (K[1]+\log (K[1]-1))}{\left (-1+e^{2 K[1]+2}\right ) K[1] \log (K[1]-1)}dK[1]\right )}{-\int _1^x\frac {\exp \left (\int _1^{K[2]}\frac {-K[1]+\log (K[1]-1)-e^{2 K[1]+2} (K[1]+\log (K[1]-1))}{\left (-1+e^{2 K[1]+2}\right ) K[1] \log (K[1]-1)}dK[1]\right ) \left (1+e^{2 K[2]+2}\right ) K[2]}{\left (-1+e^{2 K[2]+2}\right ) \log (K[2]-1)}dK[2]+c_1} \\ y(x)\to 0 \\ y(x)\to -\frac {\exp \left (\int _1^x\frac {-K[1]+\log (K[1]-1)-e^{2 K[1]+2} (K[1]+\log (K[1]-1))}{\left (-1+e^{2 K[1]+2}\right ) K[1] \log (K[1]-1)}dK[1]\right )}{\int _1^x\frac {\exp \left (\int _1^{K[2]}\frac {-K[1]+\log (K[1]-1)-e^{2 K[1]+2} (K[1]+\log (K[1]-1))}{\left (-1+e^{2 K[1]+2}\right ) K[1] \log (K[1]-1)}dK[1]\right ) \left (1+e^{2 K[2]+2}\right ) K[2]}{\left (-1+e^{2 K[2]+2}\right ) \log (K[2]-1)}dK[2]} \\ \end{align*}