Question

The complete solution of the ordinary differential equation $\frac{d^{2}y}{d{x}^2}+p\frac{dy}{dx}+qy=0$ is $y=c_1{e}^{-x}+c_2{e}^{-3x}$

Which of the following is a solution of the differential equation
$\frac{d^{2}y}{d{x}^2}+p\frac{dy}{dx}+(q+1)y=0$?

• A. $e^{-3x}$
• B. $x{e}^{-x}$
• C. $x{e}^{-2x}$
• D. $x^2{e}^{-2x}$

Answer 1

The correct option is C $x{e}^{-2x}$

$\frac{d^{2}y}{d{x}^2}+p\frac{dy}{dx}+qy=0$ ... (1)
It auxiliary equation is
$m^2+pm+q=0$ .... (2)
Since, its solution is given as
$y=c_1{e}^{-x}+c_2{e}^{-3x}$
$\Rightarrow m=-1,-3$
Hence from equation (2),
$(-1{)}^2+p(-1)+q=0$
$p-q=1$ .... (3)
and $(-3{)}^2+p(-3)+q=0$
$3p-q=9$ ... (4)
Solving (3) and (4), we get
$p=4$
$q=3$
$\because \frac{d^{2}y}{d{x}^2}+p\frac{dy}{dx}+(q+1)y=0$
Putting the value of $p$ and $q$, we get
$\frac{d^{2}y}{d{x}^2}+4\frac{dy}{dx}+4y=0$
$\therefore$ Its auxiliary equation is
$(m^2+4m+4)=0$
$m=-2,-2$
$y=(c_1+c_{2}x)e^{-2x}$
$\therefore$ Its solution is $x{e}^{-2x}$