60.1.77 problem 77

Internal problem ID [10091]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, linear first order
Problem number : 77
Date solved : Monday, January 27, 2025 at 06:26:06 PM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

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

✓ Solution by Maple

Time used: 0.046 (sec). Leaf size: 54

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

\[ y = \frac {-b x -2 \arctan \left (\frac {\tanh \left (\frac {\sqrt {a^{2}-b^{2}}\, \left (-x +c_{1} \right )}{2}\right ) \sqrt {a^{2}-b^{2}}}{a -b}\right )}{a} \]

✓ Solution by Mathematica

Time used: 0.498 (sec). Leaf size: 240

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

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