60.2.35 problem 611

Internal problem ID [10622]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, Additional non-linear first order
Problem number : 611
Date solved : Tuesday, January 28, 2025 at 04:56:15 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(y)]`]]

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

✓ Solution by Maple

Time used: 0.020 (sec). Leaf size: 44

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

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

✓ Solution by Mathematica

Time used: 0.280 (sec). Leaf size: 191

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

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