Question

The general solution of the equation $\frac{d^{3}y}{d{x}^3}+y=0$ is

• A. $c{e}^{-x}+e^{x/2}(acos\frac{\sqrt{3}}{2}x+b\sin \frac{\sqrt{3}}{2}x)$
• B. $c_1{e}^{\sqrt{3x}/2}+c_2{e}^{x/2}+c_3{e}^{-x}$
• C. $c_1{e}^{-x}+c_2{e}^{\omega x}+c_3{e}^{{\omega }^{2}x}$,where $\omega =\frac{-1+\sqrt{3}i}{2}$
• D. none of these

Answer 1

Answer: (A)

$c{e}^{-x}+e^{x/2}(acos\frac{\sqrt{3}}{2}x+b\sin \frac{\sqrt{3}}{2}x)$

$\frac{d^{3}y}{d{x}^3}+y=0$
Substitute $y=k{e}^{mx}$
$m^3+1=0$
$\Rightarrow m^3=-1$
The roots are $-1,e^{\iota \pi /3},e^{-\iota \pi /3}$
Thus the solution is
$y=c{e}^{-x}+e^{x/2}(acos\frac{\sqrt{3}}{2}x+b\sin \frac{\sqrt{3}}{2}x)$