Internal problem ID [9127]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 792.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [_Bernoulli]
\[ \boxed {y^{\prime }-\frac {y \left (-\cosh \left (\frac {1}{x +1}\right ) x +\cosh \left (\frac {1}{x +1}\right )-x +y x^{2}-x^{2}+y x^{3}\right )}{x \left (x -1\right ) \cosh \left (\frac {1}{x +1}\right )}=0} \]
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 87
dsolve(diff(y(x),x) = y(x)*(-cosh(1/(x+1))*x+cosh(1/(x+1))-x+x^2*y(x)-x^2+x^3*y(x))/x/(x-1)/cosh(1/(x+1)),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{-\left (\int \frac {\left (x^{2}+x \right ) \operatorname {sech}\left (\frac {1}{x +1}\right )+x -1}{x \left (x -1\right )}d x \right )}}{-\left (\int \frac {\operatorname {sech}\left (\frac {1}{x +1}\right ) {\mathrm e}^{-\left (\int \frac {\left (x^{2}+x \right ) \operatorname {sech}\left (\frac {1}{x +1}\right )+x -1}{x \left (x -1\right )}d x \right )} x \left (x +1\right )}{x -1}d x \right )+c_{1}} \]
✓ Solution by Mathematica
Time used: 5.482 (sec). Leaf size: 238
DSolve[y'[x] == (Sech[(1 + x)^(-1)]*y[x]*(-x - x^2 + Cosh[(1 + x)^(-1)] - x*Cosh[(1 + x)^(-1)] + x^2*y[x] + x^3*y[x]))/((-1 + x)*x),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\exp \left (\int _1^x-\frac {(K[1]+1) \text {sech}\left (\frac {1}{K[1]+1}\right ) K[1]+K[1]-1}{(K[1]-1) K[1]}dK[1]\right )}{-\int _1^x\frac {\exp \left (\int _1^{K[2]}-\frac {(K[1]+1) \text {sech}\left (\frac {1}{K[1]+1}\right ) K[1]+K[1]-1}{(K[1]-1) K[1]}dK[1]\right ) K[2] (K[2]+1) \text {sech}\left (\frac {1}{K[2]+1}\right )}{K[2]-1}dK[2]+c_1} \\ y(x)\to 0 \\ y(x)\to -\frac {\exp \left (\int _1^x-\frac {(K[1]+1) \text {sech}\left (\frac {1}{K[1]+1}\right ) K[1]+K[1]-1}{(K[1]-1) K[1]}dK[1]\right )}{\int _1^x\frac {\exp \left (\int _1^{K[2]}-\frac {(K[1]+1) \text {sech}\left (\frac {1}{K[1]+1}\right ) K[1]+K[1]-1}{(K[1]-1) K[1]}dK[1]\right ) K[2] (K[2]+1) \text {sech}\left (\frac {1}{K[2]+1}\right )}{K[2]-1}dK[2]} \\ \end{align*}