Internal problem ID [8421]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 84.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [[_homogeneous, `class C`], _dAlembert]
\[ \boxed {y^{\prime }-f \left (x a +b y\right )=0} \]
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 37
dsolve(diff(y(x),x) - f(a*x + b*y(x))=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {RootOf}\left (\left (\int _{}^{\textit {\_Z}}\frac {1}{f \left (\textit {\_a} b \right ) b +a}d \textit {\_a} \right ) b -x +c_{1} \right ) b -a x}{b} \]
✓ Solution by Mathematica
Time used: 0.211 (sec). Leaf size: 248
DSolve[y'[x] - f[a*x + b*y[x]]==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}-\frac {f(a x+b K[2]) \int _1^x\left (\frac {b^2 f'(a K[1]+b K[2])}{a+b f(a K[1]+b K[2])}-\frac {b^3 f(a K[1]+b K[2]) f'(a K[1]+b K[2])}{(a+b f(a K[1]+b K[2]))^2}\right )dK[1] b+b+a \int _1^x\left (\frac {b^2 f'(a K[1]+b K[2])}{a+b f(a K[1]+b K[2])}-\frac {b^3 f(a K[1]+b K[2]) f'(a K[1]+b K[2])}{(a+b f(a K[1]+b K[2]))^2}\right )dK[1]}{a+b f(a x+b K[2])}dK[2]+\int _1^x\frac {b f(a K[1]+b y(x))}{a+b f(a K[1]+b y(x))}dK[1]=c_1,y(x)\right ] \]