Question

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


Then, $p$ and $q$ are

• A. $p=3$, $q=3$
• B. $p=3,q=4$
• C. $p=4$,$q=3$
• D. $p=4$,$q=4$

Answer 1

The correct option is C $p=4$,$q=3$

$\frac{d^{2}y}{d{x}^2}+p\frac{dy}{dx}+qy=0$ ... (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
$p=4$
$q=3$