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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

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

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

$$x=2, y=1 / 4$$

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


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

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