A solution of y = 2xleft( frac{dy}{dx} right) | vedclass

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Question

(D) Let $P = \frac{dy}{dx}$. The given equation is $y = 2xP + x^2P^4$.Differentiating with respect to $x$:$\frac{dy}{dx} = P = 2P + 2x\frac{dP}{dx} +

Vedclass · Accepted Answer

$y = 2\sqrt{Cx} + C^2$

Vedclass · Answer

(D) Let $P = \frac{dy}{dx}$. The given equation is $y = 2xP + x^2P^4$.Differentiating with respect to $x$:$\frac{dy}{dx} = P = 2P + 2x\frac{dP}{dx} + 2xP^4 + 4x^2P^3\frac{dP}{dx}$.Rearranging the terms:$P - 2P = 2x\frac{dP}{dx} + 2xP^4 + 4x^2P^3\frac{dP}{dx}$$-P = 2x\frac{dP}{dx}(1 + 2xP^3) + 2xP^4$.Wait,let's re-evaluate the differentiation:$P = 2P + 2x\frac{dP}{dx} + 2xP^4 + 4x^2P^3\frac{dP}{dx}$$-P = 2x\frac{dP}{dx}(1 + 2xP^3) + 2xP^4$.Actually,the standard Clairaut-like form substitution $P = \sqrt{C/x}$ works directly.Substituting $P = \sqrt{C/x}$ into $y = 2xP + x^2P^4$:$y = 2x\sqrt{\frac{C}{x}} + x^2\left(\sqrt{\frac{C}{x}}\right)^4$$y = 2\sqrt{Cx} + x^2\left(\frac{C^2}{x^2}\right)$$y = 2\sqrt{Cx} + C^2$.Thus,option $D$ is correct.

$A$ solution of $y = 2x\left( \frac{dy}{dx} \right) + x^2\left( \frac{dy}{dx} \right)^4$ is

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