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$x \frac{d y}{d x}-(1+x) y=x y^{2}$

$\frac{d y}{d x}+p(x) y=f(x) y^{n}$

$\frac{d y}{d x}-\left(\frac{1}{x}+1\right) y=y^{2}$

$n=2 \quad u=y^{1-n}=y^{1-2}=y^{-1}$

$u=y^{-1}, y=u^{-1}$

$\frac{d y}{d x}=-u^{-2} \frac{d u}{d x}$

$-u^{-2} \frac{d u}{d x}-\left(\frac{1}{x}+1\right) u^{-1}=u^{-2}$

$\div-u^{-2}$

$\frac{d u}{d x}+\left(\frac{1}{x}+1\right) u=-1$

Integrating factor

$\frac{d y}{d x}+p(x) y=Q(x)$

$I F=e^{\int p(x) d x}$

$I F=e^{\int\left(\frac{1}{x}+1\right) d x}=e^{\ln x+x}=e^{x} e^{\ln x}$

$I F=x e^{x}$

$x e^{x} \frac{d u}{d x}+\left(e^{x}+x e^{x}\right) u=-x e^{x}$

$\frac{d}{d x}[I F \cdot u]=-x e^{x}$

$\int \frac{d}{d x}[{IF} \cdot u]=\int -x e^{x}$

IF $u=\int-x e^{x} d x$

$x e^{x} \cdot u=-\left[u v-\int v d u\right]=-\left[x e^{x}-\int e^{x}\right] dx$

$x e^{x} \cdot u=-x e^{x}+e^{x}+C$

$u=y^{-1}$

$x e^{x} y^{-1}=-x e^{x}+e^{x}+c \quad \div x e^{x}$

$y^{-1}=-1+\frac{1}{x}+\frac{c}{x e^{x}}$

$y^{1 / 2} \frac{d y}{d x}+y^{3 / 2}=1 \quad, y(0)=4$

$\frac{d y}{d x}+p\left(x\right) y=f(x) y^{n}$

$\frac{d y}{d x}+y=y^{-1 / 2} \quad n=-1 / 2$

$1-n=\frac{3}{2}$

$u=y^{3 / 2}, y=u^{2 / 3}, \frac{d y}{d x}=\frac{2}{3} u^{-1 / 3} \frac{d u}{d x}$

$\frac{2}{3} u^{-1 / 3} \frac{d u}{d x}+u^{2 / 3}=\left(u^{2 / 3}\right)^{-1 / 2}=u^{-1 / 3}$

$\frac{d u}{d x}+\frac{3}{2} u=\frac{3}{2}$

$\frac{d u}{d x}=\frac{3}{2}-\frac{3}{2} u=\frac{3}{2}(1-u)$

$\frac{d u}{1-u}=\frac{3}{2} d x$

$\frac{-d u}{1-u}=\frac{-3}{2} d x$

$\ln |1-u|=\frac{-3}{2} x+C$

$1-u=e^{-3 x / 2}+e^{c}$

$e^{-3 x/2+c}$

$e^{-3 x / 2} \cdot e^{c}$

$u=1-e^{-3 x / 2} e^{c}$

$u=y^{3 / 2}$

$y^{3 / 2}=1-e^{-3 x / 2} e^{c}$

$c_{1}=-e^{c}$

$y^{3 / 2}=1+e^{-3 x / 2} c_{1}$

$y(0)=4$

$8=1+c_{1}$

$c_{1}=7$

$y^{3 / 2}=1+7 e^{-3 x / 2}$