Question

The function $f\left(x\right)$ satisfying the equation $f^2\left(x\right)+4f^{\prime }\left(x\right).f\left(x\right)+\left[f^{\prime }\right(x){]}^2=0$ is-

• A. $f\left(x\right)=c.e^{(2-\sqrt{3}x)}$
• B. $f\left(x\right)=c.e^{-(2-\sqrt{3}x)}$
• C. $f\left(x\right)=c.e^{(\sqrt{3}-2)x}$
• D. $f\left(x\right)=c.e^{-(2+\sqrt{3})x}$

Answer 1

Answer: (B), (D)

The correct options are
B $f\left(x\right)=c.e^{-(2-\sqrt{3}x)}$
D $f\left(x\right)=c.e^{-(2+\sqrt{3})x}$

Given,

$f^2\left(x\right)+4f^{\prime }\left(x\right).f\left(x\right)+\left[f^{\prime }\right(x){]}^2=0$

or, $f^{\prime }\left(x\right)=\frac{-4\pm \sqrt{12}}{2}f\left(x\right)$

or, $\frac{d\left\{f\right(x\left)\right\}}{f\left(x\right)}=(-2\pm \sqrt{3})dx$

Now integrating both sides we get,

$f\left(x\right)=c{e}^{(-2\pm \sqrt{3})x}$. [ Where $c$ is integrating constant]