60.1.84 problem 84

Internal problem ID [10098]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, linear first order
Problem number : 84
Date solved : Monday, January 27, 2025 at 06:27:23 PM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

\begin{align*} y^{\prime }-f \left (a x +b y\right )&=0 \end{align*}

✓ Solution by Maple

Time used: 0.039 (sec). Leaf size: 37

dsolve(diff(y(x),x) - f(a*x + b*y(x))=0,y(x), singsol=all)
 

\[ y = \frac {\operatorname {RootOf}\left (\left (\int _{}^{\textit {\_Z}}\frac {1}{f \left (b \textit {\_a} \right ) b +a}d \textit {\_a} \right ) b -x +c_{1} \right ) b -a x}{b} \]

✓ Solution by Mathematica

Time used: 0.137 (sec). Leaf size: 248

DSolve[D[y[x],x] - f[a*x + b*y[x]]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [\int _1^{y(x)}-\frac {f(a x+b K[2]) \int _1^x\left (\frac {b^2 f''(a K[1]+b K[2])}{a+b f(a K[1]+b K[2])}-\frac {b^3 f(a K[1]+b K[2]) f''(a K[1]+b K[2])}{(a+b f(a K[1]+b K[2]))^2}\right )dK[1] b+b+a \int _1^x\left (\frac {b^2 f''(a K[1]+b K[2])}{a+b f(a K[1]+b K[2])}-\frac {b^3 f(a K[1]+b K[2]) f''(a K[1]+b K[2])}{(a+b f(a K[1]+b K[2]))^2}\right )dK[1]}{a+b f(a x+b K[2])}dK[2]+\int _1^x\frac {b f(a K[1]+b y(x))}{a+b f(a K[1]+b y(x))}dK[1]=c_1,y(x)\right ] \]