The general solution of the differential equa | vedclass

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Question

(C) Given the differential equation: $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$Simplify the numerator by factoring out $3e^{2x}$: $3e^{2

Vedclass · Accepted Answer

$y = e^{3x} + c$,where $c$ is a constant of integration.

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$Simplify the numerator by factoring out $3e^{2x}$: $3e^{2x}(1 + e^{2x})$Simplify the denominator: $e^x + e^{-x} = e^x + \frac{1}{e^x} = \frac{e^{2x} + 1}{e^x}$Substitute these back into the equation: $\frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{\frac{e^{2x} + 1}{e^x}}$Cancel the common term $(1 + e^{2x})$: $\frac{dy}{dx} = 3e^{2x} \cdot e^x = 3e^{3x}$Integrate both sides with respect to $x$: $\int dy = \int 3e^{3x} dx$$y = 3 \cdot \frac{e^{3x}}{3} + c = e^{3x} + c$

The general solution of the differential equation $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$ is

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