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Basically I tried to group the terms ydx + xdy is a perfect differential but I couldn't think a method for ($x^3$y)dx + (4x$y^4$)dy . Also dividing by xy didn't help me.

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2 Answers 2

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$$(y+ x^3y + 2x^2)\text{d}x + (x + 4xy^4+ 8y^3)\text{d}y = 0$$

$$\Rightarrow (y\text{d}x + x\text{d}y)+ x^3y\text{d}x + (2x^2\text{d}x) + 4xy^4\text{d}y+ (8y^3\text{d}y) = 0$$

$$\Rightarrow \text{d}(xy)+ (x^3y\text{d}x + 4xy^4\text{d}y) + \frac{2}{3}\text{d}(x^3) + 2\text{d}(y^4) = 0$$

$$\Rightarrow \text{d}(xy)+ xy(x^2\text{d}x + 4y^3\text{d}y) + \frac{2}{3}\text{d}(x^3) + 2\text{d}(y^4) = 0$$

$$\Rightarrow \text{d}(xy)+ xy(\frac{\text{d}(x^3)}3 + \text{d}(y^4)) + \frac{2}{3}\text{d}(x^3) + 2\text{d}(y^4) = 0$$

$$\Rightarrow \text{d}(xy)+ \frac{(xy+2)}3\text{d}(x^3) + (xy+2)\text{d}(y^4) = 0$$

$$\Rightarrow \text{d}(xy)+ (xy+2)(\frac{1}3\text{d}(x^3) + \text{d}(y^4)) = 0$$

$$\Rightarrow \frac{\text{d}(xy)}{(xy+2)}+ \frac{1}3\text{d}(x^3) + \text{d}(y^4) = 0$$

$$\Rightarrow \ln(xy+2)+ \frac{1}3x^3 + y^4 = C$$

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The implicit solution is given by Wolfram Alpha:

$$ \frac13 x^3 + y^4 + \log(x y + 2) = c $$

Since this cannot be solved for $y$, I guess further progress will be difficult.

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