<s>[INST] Find the solutions of the differential equation $y\frac{dy}{dx}+\sqrt{x-xy^4}=0$ Find the solutions of the differential equation $y\frac{dy}{dx}+\sqrt{x-xy^4}=0$
Attempt: If $u=x-xy^4$ then $\frac{du}{dx}=1-(y^4+4xy^3\frac{dy}{dx})$ But I couldn't solve it from here. Do you have any suggestion? Thanks
 [/INST] $$ y\frac{dy}{dx} + \sqrt x \sqrt{1-y^4} = 0 \\ y \frac{dy}{dx}=-\sqrt x \sqrt{1-y^4}\\ \int\frac{y}{\sqrt{1-y^4}}dy = - \int \sqrt x dx = -\frac23 x \sqrt x + C \\ \int \frac12\frac{d}{dy} \arcsin(y^2)dy = -\frac{2}{3}x\sqrt x + C \\  \arcsin(y^2) = -\frac43 x\sqrt x + C \\ y = \pm \sqrt{\sin\left(-\frac43 x\sqrt x + C \right)}$$
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