#### 2.77   ODE No. 77

$y'(x)-\cos (a y(x)+b x)=0$ Mathematica : cpu = 0.295221 (sec), leaf count = 124

$\left \{\left \{y(x)\to \frac {-2 \tan ^{-1}\left (\frac {a \tanh \left (\frac {1}{2} \left (c_1 \sqrt {a^2-b^2}-x \sqrt {a^2-b^2}\right )\right )}{\sqrt {a^2-b^2}}+\frac {b \tanh \left (\frac {1}{2} \left (c_1 \sqrt {a^2-b^2}-x \sqrt {a^2-b^2}\right )\right )}{\sqrt {a^2-b^2}}\right )-b x}{a}\right \}\right \}$ Maple : cpu = 0.052 (sec), leaf count = 54

$\left \{ y \left ( x \right ) ={\frac {1}{a} \left ( -bx+2\,\arctan \left ( {\frac {\tanh \left ( 1/2\,\sqrt {{a}^{2}-{b}^{2}} \left ( x-{\it \_C1} \right ) \right ) \sqrt {{a}^{2}-{b}^{2}}}{a-b}} \right ) \right ) } \right \}$

Hand solution

$y^{\prime }=\cos \left ( ay+bx\right )$

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);
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