\[ y'(x)-\cos (a y(x)+b x)=0 \] ✓ Mathematica : cpu = 0.317917 (sec), leaf count = 58
\[\left \{\left \{y(x)\to \frac {2 \tan ^{-1}\left (\frac {(a+b) \tanh \left (\frac {1}{2} \sqrt {a^2-b^2} \left (x-c_1\right )\right )}{\sqrt {a^2-b^2}}\right )-b x}{a}\right \}\right \}\]
✓ Maple : cpu = 0.08 (sec), leaf count = 54
\[ \left \{ y \left ( x \right ) ={\frac {1}{a} \left ( -bx+2\,\arctan \left ( {\frac {\tanh \left ( 1/2\,\sqrt {{a}^{2}-{b}^{2}} \left ( x-{\it \_C1} \right ) \right ) \sqrt {{a}^{2}-{b}^{2}}}{a-b}} \right ) \right ) } \right \} \]
\[ y^{\prime }=\cos \left ( ay+bx\right ) \]
This is separable after transformation of \(u=ay+bx\), hence \(u^{\prime }=ay^{\prime }+b\) or \(y^{\prime }=\frac {1}{a}\left ( u^{\prime }-b\right ) \). Therefore the above becomes
\begin {align*} \frac {1}{a}\left ( u^{\prime }-b\right ) & =\cos \left ( u\right ) \\ u^{\prime } & =a\cos u+b\\ \frac {du}{a\cos u+b} & =dx \end {align*}
This is the same as Kamke 76 (the problem before this), which we solved using half angle tan transformation, and the answer is
\[ u=2\arctan \left ( \frac {a+b}{\sqrt {b^{2}-a^{2}}}\tan \left ( \frac {1}{2}\sqrt {b^{2}-a^{2}}\left ( x+C\right ) \right ) \right ) \]
Since \(u=ay+bx\) then \(y=\frac {u-bx}{a}\), hence
\[ y=\frac {1}{a}\left ( 2\arctan \left ( \frac {a+b}{\sqrt {b^{2}-a^{2}}}\tan \left ( \frac {1}{2}\sqrt {b^{2}-a^{2}}\left ( x+C\right ) \right ) \right ) -bx\right ) \]
Verification
ode:=diff(y(x),x)=cos(a*y(x)+b*x); my_sol:=(1/a)*(2*arctan( (a+b)/sqrt(b^2-a^2) * tan(1/2*sqrt(b^2-a^2)*(x+_C1)))-b*x); odetest(y(x)=my_sol,ode); 0