2.32   ODE No. 32

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

$$y^{\prime }(x)+y(x{)}^2\sin (x)-2\tan (x)\sec (x)=0$$ ✓ Mathematica : cpu = 0.299935 (sec), leaf count = 34

$$\left\{\{y(x)\to \frac{\csc (x)(-2\sin (x)\cos (x)+c_1\tan (x)\sec (x))}{{\cos }^2(x)+c_1\sec (x)}\}\right\}$$ ✓ Maple : cpu = 0.239 (sec), leaf count = 28

$$\{y\left(x\right)=\frac{-2{(\cos \left(x\right))}^3\mathit{\_}\mathit{C}\mathit{1}-2}{({(\cos \left(x\right))}^3\mathit{\_}\mathit{C}\mathit{1}-2)\cos \left(x\right)}\}$$

Hand solution

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

This is Ricatti first order non-linear ODE. $P\left(x\right)=2\frac{\sin x}{{\cos }^{2}x},Q\left(x\right)=0,R\left(x\right)=-\sin \left(x\right)$. A particular solution is $y_p=\frac{1}{\cos x}$, therefore the solution is$$\begin{array}{rl}y& =y_p+\frac{1}{u}\\ y& =\frac{1}{\cos x}+\frac{1}{u}\end{array}$$

Hence$$y^{\prime }=\frac{\sin x}{{\cos }^{2}x}-\frac{u^{\prime }}{u^2}$$ Equating this to RHS of (1) gives$$\begin{array}{rl}\frac{\sin x}{{\cos }^{2}x}-\frac{u^{\prime }}{u^2}& =2\frac{\sin x}{{\cos }^{2}x}-y^2\sin \left(x\right)\\ & =2\frac{\sin x}{{\cos }^{2}x}-{(\frac{1}{\cos x}+\frac{1}{u})}^2\sin \left(x\right)\\ & =2\frac{\sin x}{{\cos }^{2}x}-(\frac{1}{{\cos }^{2}x}+\frac{1}{u^2}+\frac{2}{u\cos x})\sin \left(x\right)\end{array}$$

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

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

Integrating factor is $e^{-2\int \tan xdx}=e^{2\ln (\cos x)}={\cos }^2\left(x\right)$. Hence the above becomes$$d(u{\cos }^{2}x)={\cos }^2\left(x\right)\sin \left(x\right)$$ Integrating both sides$$\begin{array}{rl}u{\cos }^{2}x& =\int {\cos }^2\left(x\right)\sin \left(x\right)dx+C\\ & =\frac{-1}{3}{\cos }^3\left(x\right)+C\end{array}$$

Hence $$u=\frac{-1}{3}\cos \left(x\right)+\frac{C}{{\cos }^{2}x}$$ Therefore$$\begin{array}{rl}y& =y_p+\frac{1}{u}\\ & =\frac{1}{\cos x}+\frac{1}{\frac{-1}{3}\cos \left(x\right)+\frac{C}{{\cos }^{2}x}}\\ & =\frac{1}{\cos x}+\frac{3{\cos }^{2}x}{3C-{\cos }^3\left(x\right)}\end{array}$$

Let $3C=C_1$$$y=\frac{1}{\cos x}+\frac{3{\cos }^{2}x}{C_2-{\cos }^3\left(x\right)}$$ Verification

restart; 
ode:=diff(y(x),x)+y(x)^2*sin(x)-2*sin(x)/cos(x)^2 = 0; 
my_sol:=1/cos(x)+ 3*cos(x)^2/(_C1-cos(x)^3); 
odetest(y(x)=my_sol,ode); 
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