60.2.27 problem 603

Internal problem ID [10614]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, Additional non-linear first order
Problem number : 603
Date solved : Tuesday, January 28, 2025 at 04:56:01 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(y)]`]]

\begin{align*} y^{\prime }&=\frac {2 F \left (y+\ln \left (2 x +1\right )\right ) x +F \left (y+\ln \left (2 x +1\right )\right )-2}{2 x +1} \end{align*}

✓ Solution by Maple

Time used: 0.022 (sec). Leaf size: 42

dsolve(diff(y(x),x) = 1/(2*x+1)*(2*F(y(x)+ln(2*x+1))*x+F(y(x)+ln(2*x+1))-2),y(x), singsol=all)
 

\begin{align*} y &= -\ln \left (2 x +1\right )+\operatorname {RootOf}\left (F \left (\textit {\_Z} \right )\right ) \\ y &= -\ln \left (2 x +1\right )+\operatorname {RootOf}\left (-x +\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )}d \textit {\_a} +c_{1} \right ) \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.317 (sec). Leaf size: 117

DSolve[D[y[x],x] == (-2 + F[Log[1 + 2*x] + y[x]] + 2*x*F[Log[1 + 2*x] + y[x]])/(1 + 2*x),y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [\int _1^{y(x)}-\frac {F(K[2]+\log (2 x+1)) \int _1^x-\frac {2 F''(K[2]+\log (2 K[1]+1))}{F(K[2]+\log (2 K[1]+1))^2 (2 K[1]+1)}dK[1]-1}{F(K[2]+\log (2 x+1))}dK[2]+\int _1^x\left (\frac {2}{F(\log (2 K[1]+1)+y(x)) (2 K[1]+1)}-1\right )dK[1]=c_1,y(x)\right ] \]