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Find a solution for $\left(1+4 x y-4 x^{2} y\right) d x+\left(x^{2}-x^{3}\right) d y=0$ when $x=2, y=1 / 4$

(1) $\div d x\quad \left(1+4 x y-4 x^{2} y\right)+\left(x^{2}-x^{3}\right) \frac{d y}{d x}=0$

(2) $x^{2}(1-x) \frac{d y}{d x}+4 x(1-x) y=-1$

(3) $\div x^{2}(1-x) \quad \frac{d y}{d x}+\frac{4}{x} \quad y=\frac{-1}{x^{2}(1-x)}$ Linear in $y$

${a \ln}|b|=\ln |b^a|$

$\mu=e^{\int \frac{4}{x} d x}=e^{4 \ln |x|}=e^{\ln x^{4}}=x^{4}$

The general solution: $\mu y=\int \mu Q(x) d x+c$

$x^{4} y=\int x^{4} \frac{-1}{x^{2}(1-x)} d x+c=\int \frac{-x^{2}}{1-x} d x+c$ Long Division

$x^{4} y=\int\left[x+1 \frac{-1}{-x+1}\right] d x+c$

$x^{4} y=\frac{x^{2}}{2}+x\ln|-x+1 |+C$

$x=2, y=1 / 4$

$(2)^{4}\left(\frac{1}{4}\right)=\frac{(2)^{2}}{2}+2 \ln |-2+1|+c$

$C=0$

$x^{4} y=\frac{x^{2}}{2}+x\ln|-x+1|$