The function f(x) satisfies the differential | vedclass

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(D) Given the differential equation: $[f'(x)]^2 + 4f(x)f'(x) + f^2(x) = 0$. Let $y = f(x)$,then $f'(x) = \frac{dy}{dx}$. The equation becomes: $(\frac

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Both $(B)$ and $(C)$

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(D) Given the differential equation: $[f'(x)]^2 + 4f(x)f'(x) + f^2(x) = 0$. Let $y = f(x)$,then $f'(x) = \frac{dy}{dx}$. The equation becomes: $(\frac{dy}{dx})^2 + 4y(\frac{dy}{dx}) + y^2 = 0$. This is a quadratic equation in $\frac{dy}{dx}$. Using the quadratic formula $\frac{dy}{dx} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $\frac{dy}{dx} = \frac{-4y \pm \sqrt{(4y)^2 - 4(1)(y^2)}}{2(1)} = \frac{-4y \pm \sqrt{16y^2 - 4y^2}}{2} = \frac{-4y \pm \sqrt{12y^2}}{2} = \frac{-4y \pm 2\sqrt{3}y}{2} = (-2 \pm \sqrt{3})y$. Case $1$: $\frac{dy}{dx} = (-2 + \sqrt{3})y$. Separating variables: $\int \frac{dy}{y} = \int (-2 + \sqrt{3}) dx$. Integrating both sides: $\ln|y| = (-2 + \sqrt{3})x + k_1 \implies y = c_1 e^{(\sqrt{3} - 2)x}$. Case $2$: $\frac{dy}{dx} = (-2 - \sqrt{3})y$. Separating variables: $\int \frac{dy}{y} = \int (-2 - \sqrt{3}) dx$. Integrating both sides: $\ln|y| = -(2 + \sqrt{3})x + k_2 \implies y = c_2 e^{-(2 + \sqrt{3})x}$. Thus,both solutions are valid. Therefore,the correct option is $(D)$.

The function $f(x)$ satisfies the differential equation $f^2(x) + 4f'(x)f(x) + [f'(x)]^2 = 0$. Find the general solution for $f(x)$,where $c$ is an arbitrary constant.

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