2.22   ODE No. 22

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

$$y^{\prime }(x)-y(x{)}^2-y(x)\sin (2x)-\cos (2x)=0$$ ✓ Mathematica : cpu = 0.434147 (sec), leaf count = 113

$$\left\{\{y(x)\to -\frac{-\sin (x){\int }_1^{\cos (x)}\frac{e^{-K[1{]}^2}}{K[1{]}^2\sqrt{K[1{]}^2-1}}dK[1]-\frac{e^{-{\cos }^2(x)}\tan (x)}{\sqrt{{\cos }^2(x)-1}}-c_1\sin (x)}{\cos (x){\int }_1^{\cos (x)}\frac{e^{-K[1{]}^2}}{K[1{]}^2\sqrt{K[1{]}^2-1}}dK[1]+c_1\cos (x)}\}\right\}$$ ✓ Maple : cpu = 0.397 (sec), leaf count = 128

$$\{y\left(x\right)=2\frac{\sin \left(2x\right)}{\sqrt{2\cos \left(2x\right)+2}}(\mathit{\_}\mathit{C}\mathit{1}(\cos \left(2x\right)+1)\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}\mathit{P}\mathit{r}\mathit{i}\mathit{m}\mathit{e}(1,1/2,-1/2,-1,\frac{7}{8},1/2\cos \left(2x\right)+1/2)+\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}(1,1/2,-1/2,-1,\frac{7}{8},1/2\cos \left(2x\right)+1/2)\mathit{\_}\mathit{C}\mathit{1}+1/2\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}\mathit{P}\mathit{r}\mathit{i}\mathit{m}\mathit{e}(1,-1/2,-1/2,-1,\frac{7}{8},1/2\cos \left(2x\right)+1/2)\sqrt{2\cos \left(2x\right)+2}){(\mathit{\_}\mathit{C}\mathit{1}\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}(1,1/2,-1/2,-1,\frac{7}{8},1/2\cos \left(2x\right)+1/2)\sqrt{2\cos \left(2x\right)+2}+\mathit{H}\mathit{e}\mathit{u}\mathit{n}\mathit{C}(1,-1/2,-1/2,-1,\frac{7}{8},1/2\cos \left(2x\right)+1/2))}^{-1}\}$$

Hand solution

$$\begin{array}{rl}{y}^{\prime }-y^2-y\sin \left(2x\right)-\cos \left(2x\right)& =0\\ \text{(1)}& y^{\prime }& =y^2+y\sin \left(2x\right)+\cos \left(2x\right)\end{array}$$

This is Riccati first order non-linear ODE of the form of the general form $y^{\prime }=P\left(x\right)+Q\left(x\right)y+R\left(x\right)y^2$ where $P\left(x\right)=\cos \left(2x\right),Q\left(x\right)=\sin \left(2x\right),R\left(x\right)=1$. It is best to first try to spot a particular solution $y_p$ and use the transformation $y=y_p+\frac{1}{u}$ otherwise we use $y=-\frac{u^{\prime }}{yR\left(x\right)}$ transformation. For this problem $$y_p=\tan \left(x\right)$$

To verify, since $y_p^{\prime }=\frac{1}{{\cos }^{2}x}$ then plugging this particular in (1) gives

$$\frac{1}{{\cos }^{2}x}-{\tan }^2\left(x\right)-\tan \left(x\right)\sin \left(2x\right)-\cos \left(2x\right)=0$$

But $\cos \left(2x\right)={\cos }^{2}x-{\sin }^{2}x$ and $\sin \left(2x\right)=2\sin x\cos x$ and $\tan \left(x\right)=\frac{\sin x}{\cos x}$ therefore the above becomes

$$\begin{array}{rl}\frac{1}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}-\frac{\sin x}{\cos x}(2\sin x\cos x)-({\cos }^{2}x-{\sin }^{2}x)& =0\\ \frac{1}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}-2{\sin }^{2}x-{\cos }^{2}x+{\sin }^{2}x& =0\\ \frac{1}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}-{\sin }^{2}x-{\cos }^{2}x& =0\\ \frac{1}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}-1& =0\\ \frac{1-{\sin }^{2}x}{{\cos }^{2}x}-1& =0\\ \frac{{\cos }^{2}x}{{\cos }^{2}x}-1& =0\\ 1-1& =0\\ 0& =0\end{array}$$

Therefore we, we can use $y=y_p+\frac{1}{u}$

$$\begin{array}{rl}y& =\tan x+\frac{1}{u}\\ y^{\prime }& =\frac{1}{{\cos }^{2}x}-\frac{u^{\prime }}{u^2}\end{array}$$

Equating this to (1) gives

$$\begin{array}{rl}-\frac{u^{\prime }}{u^2}& =y^2+y\sin \left(2x\right)+\cos \left(2x\right)\\ -\frac{u^{\prime }}{u^2}& =-\frac{1}{{\cos }^{2}x}+{(\tan x+\frac{1}{u})}^2+(\tan x+\frac{1}{u})\sin \left(2x\right)+\cos \left(2x\right)\end{array}$$

Using $\sin \left(2x\right)=2\sin x\cos x$ and $\cos 2x={\cos }^{2}x-{\sin }^{2}x$ then above becomes

$$\begin{array}{rl}-\frac{u^{\prime }}{u^2}& =-\frac{1}{{\cos }^{2}x}+({\tan }^{2}x+\frac{1}{u^2}+\frac{2}{u}\tan x)+(\frac{\sin x}{\cos x}+\frac{1}{u})2\sin x\cos x+({\cos }^{2}x-{\sin }^{2}x)\\ u^{\prime }& =\frac{u^2}{{\cos }^{2}x}-(u^2\frac{{\sin }^{2}x}{{\cos }^{2}x}+1+2u\frac{\sin x}{\cos x})-(u^2\frac{\sin x}{\cos x}+u)2\sin x\cos x-u^2{\cos }^{2}x+u^2{\sin }^{2}x\\ & =\frac{u^2}{{\cos }^{2}x}-u^2\frac{{\sin }^{2}x}{{\cos }^{2}x}-1-2u\frac{\sin x}{\cos x}-2u^2\frac{\sin x}{\cos x}\sin x\cos x-2u\sin x\cos x-u^2{\cos }^{2}x+u^2{\sin }^{2}x\\ & =\frac{u^2}{{\cos }^{2}x}-u^2\frac{{\sin }^{2}x}{{\cos }^{2}x}-1-2u\frac{\sin x}{\cos x}-2u^2{\sin }^{2}x-2u\sin x\cos x-u^2{\cos }^{2}x+u^2{\sin }^{2}x\\ & =\frac{u^2}{{\cos }^{2}x}-u^2\frac{{\sin }^{2}x}{{\cos }^{2}x}-1-2u\frac{\sin x}{\cos x}-u^2{\sin }^{2}x-2u\sin x\cos x-u^2{\cos }^{2}x\\ & =u^2(\frac{1}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}-({\sin }^{2}x+{\cos }^{2}x))-1+u(-2\frac{\sin x}{\cos x}-2\sin x\cos x)\\ & =u^2(\frac{1-{\sin }^{2}x}{{\cos }^{2}x}-1)-1+u(-2\frac{\sin x}{\cos x}-2\sin x\cos x)\\ & =u^2(\frac{{\cos }^{2}x}{{\cos }^{2}x}-1)-1+u(-2\frac{\sin x}{\cos x}-2\sin x\cos x)\\ & =-1+2u(-\frac{\sin x}{\cos x}-\sin x\cos x)\end{array}$$

Hence

$$u^{\prime }+2u(\tan x+\sin x\cos x)=-1$$

Integrating factor is $e^{2\int \tan x+\sin x\cos xdx}$. But $$\int \tan xdx=-\ln (\cos x)$$ And $$\int \sin x\cos xdx=\frac{-1}{2}{\cos }^{2}x$$ Hence $\mu =e^{-2\ln \cos x}{e}^{-{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}{e}^{-{\cos }^{2}x}$, therefore

$$d\left(\frac{1}{{\cos }^{2}x}{e}^{-{\cos }^{2}x}u\right)=\frac{-1}{{\cos }^{2}x}{e}^{-{\cos }^{2}x}$$

Integrating both sides

$$\begin{array}{rl}\frac{1}{{\cos }^{2}x}{e}^{-{\cos }^{2}x}u& =-{\displaystyle \int }\frac{e^{-{\cos }^{2}x}}{{\cos }^{2}x}dx+C\\ u& ={\cos }^{2}x{e}^{{\cos }^{2}x}(C-{\displaystyle \int }\frac{e^{-{\cos }^{2}x}}{{\cos }^{2}x}dx)\end{array}$$

Since $y=\tan x+\frac{1}{u}$ then

$$\begin{array}{rl}y& =\tan x+\frac{1}{{\cos }^{2}x{e}^{{\cos }^{2}x}(C-{\displaystyle \int }\frac{e^{-{\cos }^{2}x}}{{\cos }^{2}x}dx)}\\ & =\tan x+\frac{e^{-{\cos }^{2}x}}{{\cos }^{2}x}{(C-{\displaystyle \int }\frac{e^{-{\cos }^{2}x}}{{\cos }^{2}x}dx)}^{-1}\end{array}$$

I do not know how Maple came up with the solution involving HeunC functions since ${\displaystyle \int }\frac{e^{-{\cos }^{2}x}}{{\cos }^{2}x}dx$ has no closed form solution. I should ask CAS experts about this.

Below is screen shot from Kamke book of the solution it gives, which matches the above result