Question

The function $f\left(x\right)$ satisfying the equation, $f^2\left(x\right)+4f^{\prime }\left(x\right).f\left(x\right)+{\left[f^{\prime }\left(x\right)\right]}^2=0,$
Where $c$ is an arbitrary constant.

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

Answer 1

Answer: (A), (B)

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

Let $f\left(x\right)=y$

and $f^{\prime }\left(x\right)=y^{\prime }$

So, $y^2+4y{y}^{\prime }+{\left(y^{\prime }\right)}^2=0$

Add $3y^2$ on both sides, we get

$4y^2+4y{y}^{\prime }+{\left(y^{\prime }\right)}^2=3y^2$

${(2y+y^{\prime })}^2=3y^2$

$2y+y^{\prime }=3y^2$

$2y+y^{\prime }=\pm \sqrt{3}y$

$(2\ne \sqrt{3})y=y^{\prime }=\frac{dy}{dx}$

$(2\ne \sqrt{3})dx=\frac{dx}{y^{\prime }}$

On inerating.

$(2\pm \sqrt{3})x=\log y+\log c$

So, $y=c{e}^{(2\pm \sqrt{3})x}$