If the differential equation begin{vmatrix} f | vedclass

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(C) The given determinant equation is $f(x)f''(x) - (f'(x))^2 = 0$.This can be rewritten as $\frac{f(x)f''(x) - (f'(x))^2}{(f(x))^2} = 0$, which is th

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$f'(x) = 2f(x)$

Vedclass · Answer

(C) The given determinant equation is $f(x)f''(x) - (f'(x))^2 = 0$.This can be rewritten as $\frac{f(x)f''(x) - (f'(x))^2}{(f(x))^2} = 0$, which is the derivative of $\frac{f'(x)}{f(x)}$.Thus, $\frac{d}{dx} \left( \frac{f'(x)}{f(x)} \right) = 0$.Integrating both sides, we get $\frac{f'(x)}{f(x)} = k$ (a constant).At $x = 0$, $k = \frac{f'(0)}{f(0)} = \frac{2}{1} = 2$.Therefore, $\frac{f'(x)}{f(x)} = 2$, which implies $f'(x) = 2f(x)$.

If the differential equation $\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$ holds for all $x$ with initial conditions $f(0) = 1$ and $f'(0) = 2$, then which of the following is true?

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