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• Notes

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 $p(0,-1)$

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