If y is a function of x,then frac{d^2y}{dx^2} | vedclass

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Question

(C) Given the differential equation $\frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0$. We know that $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$. Differentiating

Vedclass · Accepted Answer

$\frac{d^2x}{dy^2} - y \left( \frac{dx}{dy} \right)^3 = 0$

Vedclass · Answer

(C) Given the differential equation $\frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0$. We know that $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$. Differentiating with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \left( \frac{dx}{dy} \right)^{-1} \right) = -\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{d}{dx} \left( \frac{dx}{dy} \right) = -\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{d^2x}{dy^2} \cdot \frac{dy}{dx}$. Substituting $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$,we get: $\frac{d^2y}{dx^2} = -\left( \frac{dx}{dy} \right)^{-2} \cdot \frac{d^2x}{dy^2} \cdot \left( \frac{dx}{dy} \right)^{-1} = -\frac{\frac{d^2x}{dy^2}}{\left( \frac{dx}{dy} \right)^3}$. Substituting this into the original equation: $-\frac{\frac{d^2x}{dy^2}}{\left( \frac{dx}{dy} \right)^3} + y \left( \frac{1}{\frac{dx}{dy}} \right) = 0$. Multiplying by $-\left( \frac{dx}{dy} \right)^3$,we get: $\frac{d^2x}{dy^2} - y \left( \frac{dx}{dy} \right)^2 = 0$.

If $y$ is a function of $x$,then $\frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0$. If $x$ is a function of $y$,then the equation becomes:

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