#### 2.32   ODE No. 32

$y'(x)+y(x)^2 \sin (x)-2 \tan (x) \sec (x)=0$ Mathematica : cpu = 0.31515 (sec), leaf count = 34

$\left \{\left \{y(x)\to \frac {\csc (x) (c_1 \tan (x) \sec (x)-2 \sin (x) \cos (x))}{c_1 \sec (x)+\cos ^2(x)}\right \}\right \}$ Maple : cpu = 0.224 (sec), leaf count = 28

$\left \{ y \left ( x \right ) ={\frac {-2\, \left ( \cos \left ( x \right ) \right ) ^{3}{\it \_C1}-2}{ \left ( \left ( \cos \left ( x \right ) \right ) ^{3}{\it \_C1}-2 \right ) \cos \left ( x \right ) }} \right \}$

Hand solution

\begin {align} y^{\prime }+y^{2}\sin \left ( x\right ) -2\frac {\sin x}{\cos ^{2}x} & =0\nonumber \\ y^{\prime } & =2\frac {\sin x}{\cos ^{2}x}-y^{2}\sin \left ( x\right ) \nonumber \\ & =P\left ( x\right ) +Q\left ( x\right ) y+R\left ( x\right ) y^{2}\tag {1} \end {align}

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 {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 \left ( x\right ) \\ & =2\frac {\sin x}{\cos ^{2}x}-\left ( \frac {1}{\cos x}+\frac {1}{u}\right ) ^{2}\sin \left ( x\right ) \\ & =2\frac {\sin x}{\cos ^{2}x}-\left ( \frac {1}{\cos ^{2}x}+\frac {1}{u^{2}}+\frac {2}{u\cos x}\right ) \sin \left ( x\right ) \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 \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 {align*}

Or\begin {align*} 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 {align*}

Integrating factor is $$e^{-2\int \tan xdx}=e^{2\ln \left ( \cos x\right ) }=\cos ^{2}\left ( x\right )$$. Hence the above becomes$d\left ( u\cos ^{2}x\right ) =\cos ^{2}\left ( x\right ) \sin \left ( x\right )$ Integrating both sides\begin {align*} 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 {align*}

Hence $u=\frac {-1}{3}\cos \left ( x\right ) +\frac {C}{\cos ^{2}x}$ Therefore\begin {align*} 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 {align*}

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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