55.1.4 problem 1.1.4

Internal problem ID [13224]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, First-Order differential equations
Problem number : 1.1.4
Date solved : Wednesday, October 01, 2025 at 03:37:16 AM
CAS classification : [_linear]

\begin{align*} g \left (x \right ) y^{\prime }&=f_{1} \left (x \right ) y+f_{0} \left (x \right ) \end{align*}
Maple. Time used: 0.002 (sec). Leaf size: 38
ode:=g(x)*diff(y(x),x) = f__1(x)*y(x)+f__0(x); 
dsolve(ode,y(x), singsol=all);
 
\[ y = \left (\int \frac {f_{0} \left (x \right ) {\mathrm e}^{-\int \frac {f_{1} \left (x \right )}{g \left (x \right )}d x}}{g \left (x \right )}d x +c_1 \right ) {\mathrm e}^{\int \frac {f_{1} \left (x \right )}{g \left (x \right )}d x} \]
Mathematica. Time used: 0.053 (sec). Leaf size: 64
ode=g[x]*D[y[x],x]==f1[x]*y[x]+f0[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
\begin{align*} y(x)&\to \exp \left (\int _1^x\frac {\text {f1}(K[1])}{g(K[1])}dK[1]\right ) \left (\int _1^x\frac {\exp \left (-\int _1^{K[2]}\frac {\text {f1}(K[1])}{g(K[1])}dK[1]\right ) \text {f0}(K[2])}{g(K[2])}dK[2]+c_1\right ) \end{align*}
Sympy. Time used: 18.388 (sec). Leaf size: 85
from sympy import * 
x = symbols("x") 
y = Function("y") 
f0 = Function("f0") 
f1 = Function("f1") 
g = Function("g") 
ode = Eq(-f0(x) - f1(x)*y(x) + g(x)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
\[ y{\left (x \right )} = \frac {\left (C_{1} + \int \frac {f_{0}{\left (x \right )} e^{- \int \frac {f_{1}{\left (x \right )}}{g{\left (x \right )}}\, dx}}{g{\left (x \right )}}\, dx + \int \frac {f_{1}{\left (x \right )} y{\left (x \right )} e^{- \int \frac {f_{1}{\left (x \right )}}{g{\left (x \right )}}\, dx}}{g{\left (x \right )}}\, dx\right ) e^{\int \frac {f_{1}{\left (x \right )}}{g{\left (x \right )}}\, dx}}{\left (e^{\int \frac {f_{1}{\left (x \right )}}{g{\left (x \right )}}\, dx}\right ) \int \frac {f_{1}{\left (x \right )} e^{- \int \frac {f_{1}{\left (x \right )}}{g{\left (x \right )}}\, dx}}{g{\left (x \right )}}\, dx + 1} \]