If f(x), f^{prime}(x), f^{prime prime}(x) are | vedclass

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(A) Given the determinant equation: $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}ight|=0$ Expandin

Vedclass · Accepted Answer

$e^{2 x}$

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(A) Given the determinant equation: $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}ight|=0$ Expanding the determinant,we get: $f(x) f^{\prime \prime}(x) - (f^{\prime}(x))^2 = 0$ $\Rightarrow f(x) f^{\prime \prime}(x) = (f^{\prime}(x))^2$ Dividing both sides by $f(x) f^{\prime}(x)$ (assuming $f(x), f^{\prime}(x) 
eq 0$): $\frac{f^{\prime \prime}(x)}{f^{\prime}(x)} = \frac{f^{\prime}(x)}{f(x)}$ Integrating both sides with respect to $x$: $\int \frac{f^{\prime \prime}(x)}{f^{\prime}(x)} dx = \int \frac{f^{\prime}(x)}{f(x)} dx$ $\ln(f^{\prime}(x)) = \ln(f(x)) + C_1$ Using initial conditions $f(0)=1$ and $f^{\prime}(0)=2$: $\ln(2) = \ln(1) + C_1 \Rightarrow C_1 = \ln(2)$ So,$\ln(f^{\prime}(x)) = \ln(f(x)) + \ln(2) = \ln(2f(x))$ $\Rightarrow f^{\prime}(x) = 2f(x)$ $\Rightarrow \frac{f^{\prime}(x)}{f(x)} = 2$ Integrating again: $\int \frac{f^{\prime}(x)}{f(x)} dx = \int 2 dx$ $\ln(f(x)) = 2x + C_2$ Using $f(0)=1$: $\ln(1) = 2(0) + C_2 \Rightarrow C_2 = 0$ Thus,$\ln(f(x)) = 2x \Rightarrow f(x) = e^{2x}$ Therefore,the correct option is $A$.

If $f(x), f^{\prime}(x), f^{\prime \prime}(x)$ are positive functions and $f(0)=1, f^{\prime}(0)=2$,then the solution of the differential equation $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$ is

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