47.2.41 problem 39

Internal problem ID [7457]
Book : Ordinary differential equations and calculus of variations. Makarets and Reshetnyak. Wold Scientific. Singapore. 1995
Section : Chapter 1. First order differential equations. Section 1.2 Homogeneous equations problems. page 12
Problem number : 39
Date solved : Monday, January 27, 2025 at 03:00:44 PM
CAS classification : [[_homogeneous, `class C`], _exact, _dAlembert]

\begin{align*} \left (y^{\prime }+1\right ) \ln \left (\frac {x +y}{x +3}\right )&=\frac {x +y}{x +3} \end{align*}

✓ Solution by Maple

Time used: 0.143 (sec). Leaf size: 40

dsolve((diff(y(x),x)+1)*ln((y(x)+x)/(x+3))=(y(x)+x)/(x+3),y(x), singsol=all)
 

\[ y = \frac {-x \operatorname {LambertW}\left (\frac {{\mathrm e}^{-1}}{\left (x +3\right ) c_{1}}\right ) c_{1} +1}{\operatorname {LambertW}\left (\frac {{\mathrm e}^{-1}}{\left (x +3\right ) c_{1}}\right ) c_{1}} \]

✓ Solution by Mathematica

Time used: 0.241 (sec). Leaf size: 30

DSolve[(D[y[x],x]+1)*Log[(y[x]+x)/(x+3)]==(y[x]+x)/(x+3),y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [-y(x)+(y(x)+x) \log \left (\frac {y(x)+x}{x+3}\right )-x=c_1,y(x)\right ] \]