The function y=e^{kx} satisfies (frac{d^2y}{d | vedclass

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(C) Given $y = e^{kx}$.Then $\frac{dy}{dx} = ke^{kx} = ky$ and $\frac{d^2y}{dx^2} = k^2e^{kx} = k^2y$.Substituting these into the given equation:$(\fr

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three distinct values of $k$

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(C) Given $y = e^{kx}$.Then $\frac{dy}{dx} = ke^{kx} = ky$ and $\frac{d^2y}{dx^2} = k^2e^{kx} = k^2y$.Substituting these into the given equation:$(\frac{d^2y}{dx^2} + \frac{dy}{dx})(\frac{dy}{dx} - y) = y\frac{dy}{dx}$$(k^2y + ky)(ky - y) = y(ky)$$ky(k+1) \cdot y(k-1) = ky^2$$k(k^2 - 1)y^2 = ky^2$Since $y = e^{kx} 
eq 0$, we can divide by $y^2$:$k(k^2 - 1) = k$$k^3 - k = k$$k^3 - 2k = 0$$k(k^2 - 2) = 0$Thus, $k = 0$ or $k^2 = 2$, which gives $k = 0, \sqrt{2}, -\sqrt{2}$.There are three distinct values of $k$.

The function $y=e^{kx}$ satisfies $(\frac{d^2y}{dx^2}+\frac{dy}{dx})(\frac{dy}{dx}-y)=y\frac{dy}{dx}$. It is valid for

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