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I'm trying to see if it is possible to obtain an explicit form of the following differential equation

$$\left(\frac{dy}{dx}\right)^{2}=\frac{1}{ay^2+by+c}$$

where $a,b$ and $c\in\mathbb{R}$\{$0$}

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Hint: $$ \begin{align} x&=\int\sqrt{ay^2+by+c}\ \,\mathrm{d}y\\ &=\sqrt{a}\int\sqrt{\left(y-\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}}\ \,\mathrm{d}y\\ \end{align} $$ A trig substitution often helps, but the particular substitution would depend on the sign of $b^2-4ac$.

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HINT

Inverse your equation and get dx/dy. You now have to find x(y) and, may be, you could later inverse again to get y(x).

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