Question

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 $\left|\begin{array}{cc}f\left(x\right)& f^{\prime }\left(x\right)\\ f^{\prime }\left(x\right)& f^{\prime \prime }\left(x\right)\end{array}\right|=0$ is

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

Answer 1

Answer: (A)

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