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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

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

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

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

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

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

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