Question

If $f(x), f'(x)- f''(x)$ are positive functions and $f(0) = 1. f'(0) = 2$, then the solution of the differential equation $\begin{vmatrix}f\left(x\right)&f'\left(x\right)\\ f'\left(x\right)&f''\left(x\right)\end{vmatrix}=0$ is

• A. $e^{2x}$
• B. $2\, sin x + 1$
• C. $\sin^2 x + 2x + 1$
• D. $e^{4x}$

Answer 1

Answer: (A)

Since, $|\begin{vmatrix}f(x) & f'(x) \\ f'(x) & f''(x)\end{vmatrix} =0$
$\Rightarrow f(x) f''(x)-\left(f'(x)\right)^{2}=0$
$\Rightarrow f(x) f''(x)=\left(f'(x)\right)^{2}$
$\Rightarrow \frac{f''(x)}{f'(x)}=\frac{f'(x)}{f(x)}$
On, integrating both sides, we get
$\int \frac{f''(x)}{f'(x)} d x=\int \frac{f'(x)}{f(x)} d x$
$\Rightarrow \log _{e}\left(f'(x)\right) =\log _{e}(f(x))+c$
$\therefore f(0) =1$ and $f'(0)=2$
$c=\log _{e} 2$
$\Rightarrow \log _{e}\left(f'(x)\right) =\log _{e}(f(x))+\log _{e} 2$
$\Rightarrow f'(x) =2 f(x)$
$\frac{f'(x)}{f(x)} =2$
$\Rightarrow \int \frac{f'(x)}{f(x)} d x=\int 2 d x$
$\Rightarrow \log _{e} f(x)=2 x +c$
$\because f(0)=1, \text { so } c=0$
$\because \log _{e} f(x)=2 x$
$\Rightarrow f(x)=e^{2 x}$