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Employ two different methods to find the general solution of the equation
$$y^{\prime}=\left(y+y^{-3}\right) \cos x$$
Then determine the solution curve that passes through the point $$P(0,-1)$$

(1) $$\frac{d y}{d x}=\left(y+y^{-3}\right) \cos x \quad * d x$$

$$d y=\left(y+y^{3}\right) \cos x d x \quad \div\left(y+y^{3}\right)$$

$$\frac{d y}{y+y^{-3}}=\cos x d x$$

(2) Integrate

$$\int \frac{d y}{y+y^{-3}}=\int \cos x d x$$

$$\int \frac{d y \cdot y^{3}}{\left(y+y^{-3}\right) y^{3}}=\int \cos x d x$$

$$\rightarrow \int \frac{y^{3} d y}{y^{4}+1}=\sin x+c$$

$$\int \frac{4 y^{3} d y}{4(y^{4}+1)}=\sin x+c$$

$$\frac{1}{4} \cdot \ln \left|y^{4}+1\right|=\sin x+c$$

at $$P(0,-1)$$

$$\rightarrow \frac{1}{4} \ln \left|(-1)^{4}+1\right|=\sin (0)+c$$

$$\longrightarrow c=\frac{1}{4} \ln (2)$$

$$\frac{d y}{d x}=\cos x(y)+\cos x\left(y^{-3}\right) \longrightarrow \frac{d y}{d x}-\cos (x)(y)=\cos x\left(y^{-3}\right) \quad * y^{3}$$

$$y^{3} \frac{d y}{d x}-\cos (x)\left(y^{4}\right)=\cos (x) \cdots-(1)$$

let $$z=y^{4} \rightarrow \frac{d z}{d x}=4 y^{3} \frac{d y}{d x} \quad \div 4$$

$$\frac{1}{4} \frac{d z}{d x}=y^{3} \frac{d y}{d x} \cdots (2)$$

$$\frac{1}{4} \frac{d z}{d x}-\cos (x) z=\cos x \quad * 4$$

$$\rightarrow \frac{d z}{d x}-4 \cos x z=4 \cos x$$ Linear in $$z$$

$$\mu=e^{\int-4 \cos x d x}=e^{-4 \sin x}$$

$$\mu \cdot z=\int e^{-4 \sin x} \cdot 4 \cos x d x$$

$$e^{-4 \sin x} \cdot y^{4}=4 \frac{1}{-4 \cos x} e^{-4 \sin x} \cos x d x=-e^{-4 \sin x}+c$$

at $$p(0,-1)$$

$$e^{-4 \sin (0)} \cdot {(-1)^{4}}=-e^{-4 \sin (0)}+c \rightarrow c=2$$

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