Question

The differential equation whose general solution is given by,
$y=\left(c_{1}cos\right(x+c_2\left)\right)-\left(c_3{e}^{(-x+c_4)}\right)+\left(c_{5}sinx\right)$ where $c_1,c_2,c_3,c_4,c_5$ are arbitrary constants, is

• A. $\frac{d^{4}y}{d{x}^4}-\frac{d^{2}y}{d{x}^2}+y=0$
• B. $\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$
• C. $\frac{d^{5}y}{d{x}^5}+y=0$
• D. $\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-y=0$

Answer 1

The correct option is B $\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$

$y=c_{1}cos(x+c_2)-\left(c_3{e}^{-x+c_4}\right)+\left(c_{5}sinx\right)$
$\Rightarrow y=c_1(cosxcos{c}_2-sinxsin{c}_2)-\left(c_3{e}^{c_4}{e}^{-x}\right)+\left(c_{5}sinx\right)$
$\Rightarrow y=\left(c_{1}cos{c}_2\right)cosx-(c_{1}sin{c}_2-c_5)sinx-\left(c_3{e}^{c_4}\right)e^{-x}$
$\Rightarrow y=1cosx+msinx-n{e}^{-x}$ (1)
Where, l, m, n are arbitrary constant
$\Rightarrow \frac{dy}{dx}=-lsinx+mcosx+n{e}^{-x}$ (2)
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-lcosx-msinx-n{e}^{-x}$ (3)
$\Rightarrow \frac{d^{3}y}{d{x}^3}=lsinx-mcosx+n{e}^{-x}$ (4)
From equations (1) +(3), $\frac{d^{2}y}{d{x}^2}+y=-2n{e}^{-x}$ (5)
From equations (2) +(4), $\frac{d^{3}y}{d{x}^3}+\frac{dy}{dx}=2n{e}^{-x}$ (6)
From equations (5) + (6), we get $\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$