\[ y'(x)+y(x)^2 \sin (x)-2 \tan (x) \sec (x)=0 \] ✓ Mathematica : cpu = 0.328093 (sec), leaf count = 34
\[\left \{\left \{y(x)\to \frac {\csc (x) (-2 \sin (x) \cos (x)+c_1 \tan (x) \sec (x))}{\cos ^2(x)+c_1 \sec (x)}\right \}\right \}\] ✓ Maple : cpu = 0.26 (sec), leaf count = 28
\[y \relax (x ) = \frac {-2 \left (\cos ^{3}\relax (x )\right ) c_{1}-2}{\left (\left (\cos ^{3}\relax (x )\right ) c_{1}-2\right ) \cos \relax (x )}\]
Hand solution
\begin {align} y^{\prime }+y^{2}\sin \relax (x) -2\frac {\sin x}{\cos ^{2}x} & =0\nonumber \\ y^{\prime } & =2\frac {\sin x}{\cos ^{2}x}-y^{2}\sin \relax (x) \nonumber \\ & =P\relax (x) +Q\relax (x) y+R\relax (x) y^{2}\tag {1} \end {align}
This is Ricatti first order non-linear ODE. \(P\relax (x) =2\frac {\sin x}{\cos ^{2}x},Q\relax (x) =0,R\relax (x) =-\sin \relax (x) \). A particular solution is \(y_{p}=\frac {1}{\cos x}\), therefore the solution is\begin {align*} y & =y_{p}+\frac {1}{u}\\ y & =\frac {1}{\cos x}+\frac {1}{u} \end {align*}
Hence\[ y^{\prime }=\frac {\sin x}{\cos ^{2}x}-\frac {u^{\prime }}{u^{2}}\] Equating this to RHS of (1) gives\begin {align*} \frac {\sin x}{\cos ^{2}x}-\frac {u^{\prime }}{u^{2}} & =2\frac {\sin x}{\cos ^{2}x}-y^{2}\sin \relax (x) \\ & =2\frac {\sin x}{\cos ^{2}x}-\left (\frac {1}{\cos x}+\frac {1}{u}\right ) ^{2}\sin \relax (x) \\ & =2\frac {\sin x}{\cos ^{2}x}-\left (\frac {1}{\cos ^{2}x}+\frac {1}{u^{2}}+\frac {2}{u\cos x}\right ) \sin \relax (x) \end {align*}
Hence\begin {align*} -\frac {u^{\prime }}{u^{2}} & =-\frac {\sin x}{\cos ^{2}x}+2\frac {\sin x}{\cos ^{2}x}-\frac {\sin \relax (x) }{\cos ^{2}x}-\frac {\sin \left ( x\right ) }{u^{2}}-\frac {2\sin \relax (x) }{u\cos x}\\ & =-\frac {\sin \relax (x) }{u^{2}}-\frac {2\sin \relax (x) }{u\cos x} \end {align*}
Or\begin {align*} u^{\prime } & =\sin \relax (x) +\frac {2u\sin \relax (x) }{\cos x}\\ u^{\prime }-2u\tan \relax (x) & =\sin \relax (x) \end {align*}
Integrating factor is \(e^{-2\int \tan xdx}=e^{2\ln \left (\cos x\right ) }=\cos ^{2}\relax (x) \). Hence the above becomes\[ d\left (u\cos ^{2}x\right ) =\cos ^{2}\relax (x) \sin \relax (x) \] Integrating both sides\begin {align*} u\cos ^{2}x & =\int \cos ^{2}\relax (x) \sin \relax (x) dx+C\\ & =\frac {-1}{3}\cos ^{3}\relax (x) +C \end {align*}
Hence \[ u=\frac {-1}{3}\cos \relax (x) +\frac {C}{\cos ^{2}x}\] Therefore\begin {align*} y & =y_{p}+\frac {1}{u}\\ & =\frac {1}{\cos x}+\frac {1}{\frac {-1}{3}\cos \relax (x) +\frac {C}{\cos ^{2}x}}\\ & =\frac {1}{\cos x}+\frac {3\cos ^{2}x}{3C-\cos ^{3}\relax (x) } \end {align*}
Let \(3C=C_{1}\)\[ y=\frac {1}{\cos x}+\frac {3\cos ^{2}x}{C_{2}-\cos ^{3}\relax (x) }\] Verification