2.77   ODE No. 77

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

$$y^{\prime }(x)-\cos (ay(x)+bx)=0$$ ✓ Mathematica : cpu = 0.277764 (sec), leaf count = 124

$$\left\{\{y(x)\to \frac{-bx-2{\tan }^{-1}(\frac{a\tanh \left(\frac{1}{2}(-x\sqrt{a^2-b^2}+c_1\sqrt{a^2-b^2})\right)}{\sqrt{a^2-b^2}}+\frac{b\tanh \left(\frac{1}{2}(-x\sqrt{a^2-b^2}+c_1\sqrt{a^2-b^2})\right)}{\sqrt{a^2-b^2}})}{a}\}\right\}$$ ✓ Maple : cpu = 0.056 (sec), leaf count = 54

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

Hand solution

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

This is separable after transformation of $u=ay+bx$, hence $u^{\prime }=a{y}^{\prime }+b$ or $y^{\prime }=\frac{1}{a}(u^{\prime }-b)$. Therefore the above becomes

$$\begin{array}{rl}\frac{1}{a}(u^{\prime }-b)& =\cos \left(u\right)\\ u^{\prime }& =a\cos u+b\\ \frac{du}{a\cos u+b}& =dx\end{array}$$

This is the same as Kamke 76 (the problem before this), which we solved using half angle tan transformation, and the answer is

$$u=2\arctan (\frac{a+b}{\sqrt{b^2-a^2}}\tan \left(\frac{1}{2}\sqrt{b^2-a^2}(x+C)\right))$$

Since $u=ay+bx$ then $y=\frac{u-bx}{a}$, hence

$$y=\frac{1}{a}(2\arctan (\frac{a+b}{\sqrt{b^2-a^2}}\tan \left(\frac{1}{2}\sqrt{b^2-a^2}(x+C)\right))-bx)$$

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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