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Solve the equation $x \sec y d x+\left(1+x^{2} \sec y \tan y\right) d y=0$

(1) $x \sec y d x+\left(1+x^{2} \sec y \tan y\right) d y=0$

(2) $\frac{d M}{d y}=\frac{d N}{d x}$

$\frac{d M}{d y}=x \sec y \tan y$

$\frac{d N}{d x}=2 x \sec y \tan y$

NoT Exact

(1) $\frac{d M}{d y}-\frac{d N}{d x}=-x \sec y \tan y \simeq M$

(2) $\frac{\frac{d M}{d y}-\frac{d N}{d x}}{x \sec y}=\frac{-x \sec y \tan y}{x \sec y}= - \tan y$

$I \cdot F=e^{-\int \tan y d y}=e^{\ln |\sec y|}=\sec y$

$\sec y\left(x \sec y d x+\left(1+x^{2} \sec y \tan y\right) d y\right)=0$

$x \sec ^{2} y d x+\sec y+\left(x^{2} \sec ^{2} y \tan y\right) d y=0$

$\psi(x, y)=\int x \sec ^{2} y d x+g(y)$

$=\frac{x^{2}}{2}{\sec ^{2} y}+g(y)$

$\frac{d \psi}{d y}=\frac{x^{2}}{2} 2 \cdot \sec ^{2} y \tan y+g^{\prime}(y)=\sec y+x^{2} \sec ^{2} y \tan y$

$g^{\prime}(y)=\sec y \Rightarrow g(y)=\ln |\sec y+\tan y|$

$\frac{x^{2}}{2} \sec ^{2} y+\ln |\sec y+\tan y|=c$

Show that $M(x, y)=y e^{x}$ is an integrating factor for the equation then solve it
$\left(\frac{\sin y}{y}-3 e^{-x} \sin x\right)+\left(\frac{\cos y+3 e^{-x} \cos x}{y}\right) y^{\prime}=0$

$\left(\frac{\sin y}{y}-3 e^{-x} \sin x\right)+\left(\frac{\cos y+3 e^{-x} \cos x}{y}\right) y^{\prime}=0 \quad \quad y e^{x}$

(1) $e^{x} \sin y-3 y \sin x+\left(e^{x} \cos y+3 \cos x\right) y^{\prime}=0$

(2) $\frac{d M}{d y}=\frac{d N}{d x}$

$\frac{d M}{d y}=e^{x} \cos y-3 \sin (x)$

$\frac{d N}{d x}=e^{x} \cos y-3 \sin (x)$

Exact

$\psi(x, y)=\int\left(e^{x} \sin y-3 y \sin x\right) d x+g(y)$

$\psi(x, y)=e^{x} \sin y+3 y \cos x+g(y)$

$\frac{d \psi}{d y}=e^{x} \cos y+3 \cos x+g^{\prime}(y)=e^{x} \cos y+3 \cos x$

$g(y)=\int 0 d y=c$

$\psi(x, y)=e^{x} \sin y+3 y \cos x=c$