The differential equation obtained by elimina | vedclass

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(B) Given equation is $y^{2}=(2 x+c)^{5}$ ...$(i)$Differentiating both sides with respect to $x$:$2 y \frac{dy}{dx} = 5(2 x+c)^{4} 	imes 2$$y \frac{d

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$\left(\frac{d y}{d x}\right)^{5}-3125 y^{3}=0$

Vedclass · Answer

(B) Given equation is $y^{2}=(2 x+c)^{5}$ ...$(i)$Differentiating both sides with respect to $x$:$2 y \frac{dy}{dx} = 5(2 x+c)^{4} 	imes 2$$y \frac{dy}{dx} = 5(2 x+c)^{4}$From this,we find $(2 x+c)$:$(2 x+c)^{4} = \frac{y}{5} \frac{dy}{dx}$$(2 x+c) = \left(\frac{y}{5} \frac{dy}{dx}ight)^{1/4}$Substituting this into equation $(i)$:$y^{2} = \left[\left(\frac{y}{5} \frac{dy}{dx}ight)^{1/4}ight]^{5}$$y^{2} = \left(\frac{y}{5} \frac{dy}{dx}ight)^{5/4}$Raising both sides to the power of $4$:$(y^{2})^{4} = \left(\frac{y}{5} \frac{dy}{dx}ight)^{5}$$y^{8} = \frac{y^{5}}{5^{5}} \left(\frac{dy}{dx}ight)^{5}$$y^{8} = \frac{y^{5}}{3125} \left(\frac{dy}{dx}ight)^{5}$Dividing by $y^{5}$ (assuming $y 
eq 0$):$y^{3} = \frac{1}{3125} \left(\frac{dy}{dx}ight)^{5}$$3125 y^{3} = \left(\frac{dy}{dx}ight)^{5}$$\left(\frac{dy}{dx}ight)^{5} - 3125 y^{3} = 0$

The differential equation obtained by eliminating the arbitrary constants from the equation $y^{2}=(2 x+c)^{5}$ is

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