2.76   ODE No. 76

1. Problem in Latex
2. Mathematica input
3. Maple input

$$-a\cos (y(x))+b+y^{\prime }(x)=0$$ ✓ Mathematica : cpu = 0.119122 (sec), leaf count = 51

$$\left\{\{y(x)\to 2{\tan }^{-1}\left(\frac{(a-b)\tanh \left(\frac{1}{2}\sqrt{a^2-b^2}(x-c_1)\right)}{\sqrt{a^2-b^2}}\right)\}\right\}$$

✓ Maple : cpu = 0.057 (sec), leaf count = 41

$$\{y\left(x\right)=2\arctan \left(\frac{\tanh \left(1/2\sqrt{a^2-b^2}(x+\mathit{\_}\mathit{C}\mathit{1})\right)\sqrt{a^2-b^2}}{a+b}\right)\}$$

Hand solution

$$y^{\prime }=a\cos y+b$$

This is separable.

$$\begin{array}{rl}\frac{dy}{a\cos y+b}& =dx\\ \text{(1)}& \int \frac{dy}{a\cos y+b}& =x+C\end{array}$$

Using standard Tangent half-angle substitution, let $t=\tan \frac{y}{2},\cos y=\frac{1-t^2}{1+t^2},dy=\frac{2}{1+t^2}dt$, then the integral becomes

$$\begin{array}{rl}\int \frac{dy}{a\cos y+b}& =\int \frac{2}{1+t^2}\frac{1}{(a\frac{1-t^2}{1+t^2}+b)}dt\\ & =2\int \frac{1+t^2}{(1+t^2)(a(1-t^2)+b(1+t^2))}dt\\ & =2\int \frac{dt}{a-a{t}^2+b+b{t}^2}\\ & =2\int \frac{dt}{(a+b)+t^2(b-a)}\\ & =2\int \frac{dt}{(a+b)(1+\frac{t^2(b-a)}{(a+b)})}\\ & =\frac{2}{a+b}\int \frac{dt}{(1+\frac{t^2(b-a)}{(a+b)})}\end{array}$$

Let $z^2=\frac{t^2(b-a)}{(a+b)}$, or $z=\frac{t\sqrt{b-a}}{\sqrt{a+b}}$, then $\frac{dz}{dt}=\frac{\sqrt{b-a}}{\sqrt{a+b}}$ and the above integral becomes

$$\begin{array}{rl}\frac{2}{a+b}\int \frac{dt}{(1+\frac{t^2(b-a)}{(a+b)})}& =\frac{2}{a+b}\int \frac{\sqrt{a+b}}{\sqrt{b-a}}\frac{dz}{(1+z^2)}\\ & =\frac{2}{a+b}\frac{\sqrt{a+b}}{\sqrt{b-a}}\int \frac{dz}{(1+z^2)}\\ & =\frac{2}{\sqrt{a+b}}\frac{1}{\sqrt{b-a}}\int \frac{dz}{(1+z^2)}\\ & =\frac{2}{\sqrt{(a+b)(b-a)}}\int \frac{dz}{(1+z^2)}\\ & =\frac{2}{\sqrt{b^2-a^2}}\int \frac{dz}{(1+z^2)}\end{array}$$

Now, $\int \frac{dz}{(1+z^2)}=\arctan \left(z\right)$, hence

$$\begin{array}{rl}\frac{2}{\sqrt{b^2-a^2}}\int \frac{dz}{(1+z^2)}& =\frac{2}{\sqrt{b^2-a^2}}\arctan \left(z\right)\\ & =\frac{2}{\sqrt{b^2-a^2}}\arctan \left(\frac{t\sqrt{b-a}}{\sqrt{a+b}}\right)\end{array}$$

But $t=\tan \frac{y}{2}$ therefore

$$\frac{2}{\sqrt{b^2-a^2}}\arctan \left(\frac{t\sqrt{b-a}}{\sqrt{a+b}}\right)=\frac{2}{\sqrt{b^2-a^2}}\arctan \left(\frac{\tan \left(\frac{y}{2}\right)\sqrt{b-a}}{\sqrt{a+b}}\right)$$

Going back to (1)

$$\begin{array}{rl}\int \frac{dy}{a\cos y+b}& =x+C\\ \frac{2}{\sqrt{b^2-a^2}}\arctan \left(\frac{\tan \left(\frac{y}{2}\right)\sqrt{b-a}}{\sqrt{a+b}}\right)& =x+C\\ \arctan \left(\frac{\tan \left(\frac{y}{2}\right)\sqrt{b-a}}{\sqrt{a+b}}\right)& =\frac{1}{2}\sqrt{b^2-a^2}(x+C)\\ \frac{\tan \left(\frac{y}{2}\right)\sqrt{b-a}}{\sqrt{a+b}}& =\tan \left(\frac{1}{2}\sqrt{b^2-a^2}(x+C)\right)\\ \tan \left(\frac{y}{2}\right)& =\frac{\sqrt{a+b}}{\sqrt{b-a}}\tan \left(\frac{1}{2}\sqrt{b^2-a^2}(x+C)\right)\\ \frac{y}{2}& =\arctan (\frac{(a+b)}{\sqrt{(a+b)(b-a)}}\tan \left(\frac{1}{2}\sqrt{b^2-a^2}(x+C)\right))\\ & =\arctan (\frac{(a+b)}{\sqrt{b^2-a^2}}\tan \left(\frac{1}{2}\sqrt{b^2-a^2}(x+C)\right))\\ y& =2\arctan (\frac{a+b}{\sqrt{b^2-a^2}}\tan \left(\frac{1}{2}\sqrt{b^2-a^2}(x+C)\right))\end{array}$$

Verification

ode:=diff(y(x),x)=a*cos(y(x))+b; 
my_sol:=2*arctan( (a+b)/sqrt(b^2-a^2) * tan(1/2*sqrt(b^2-a^2)*(x+_C1))); 
odetest(y(x)=my_sol,ode); 
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