The differential equation of y = A{e^{2x}} + | vedclass

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(D) Given the equation: $y = Ae^{2x} + Be^{-2x}$ Differentiating with respect to $x$ once: $\frac{dy}{dx} = 2Ae^{2x} - 2Be^{-2x}$ Differentiating with

Vedclass · Accepted Answer

$\frac{d^2y}{dx^2} = 4y$

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(D) Given the equation: $y = Ae^{2x} + Be^{-2x}$ Differentiating with respect to $x$ once: $\frac{dy}{dx} = 2Ae^{2x} - 2Be^{-2x}$ Differentiating with respect to $x$ again: $\frac{d^2y}{dx^2} = 4Ae^{2x} + 4Be^{-2x}$ Factoring out $4$: $\frac{d^2y}{dx^2} = 4(Ae^{2x} + Be^{-2x})$ Since $y = Ae^{2x} + Be^{-2x}$,we substitute $y$ into the equation: $\frac{d^2y}{dx^2} = 4y$

The differential equation of $y = A{e^{2x}} + B{e^{ - 2x}}$ is (where $A$ and $B$ are arbitrary constants):

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