Question

The differential equation whose general solution is given by, $y=\left(c_1\cos \right(x+c_2\left)\right)-\left(c_3{e}^{(-x+c_4)}\right)+\left(c_5\sin x\right),$ where $c_1,c_2,c_3,c_4,c_5$ are 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 $\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$

$y=c_1\cos (x+c_2)-\left(c_3{e}^{-x+c_4}\right)+\left(c_5\sin x\right)$
$\Rightarrow y=c_1(\cos x\cos c_2-\sin x\sin c_2)-\left(c_3{e}^{c_4}{e}^{-x}\right)+\left(c_5\sin x\right)$
$\Rightarrow y=\left(c_1\cos c_2\right)\cos x-(c_1\sin c_2-c_5)\sin x-\left(c_3{e}^{c_4}\right)e^{-x}$
$\Rightarrow y=l\cos x+m\sin x-n{e}^{-x}\cdots \left(i\right)$
where $l,m,n$ are arbitrary constant
$\Rightarrow \frac{dy}{dx}=-l\sin x+m\cos x+n{e}^{-x}\cdots \left(ii\right)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-l\cos x-m\sin x-n{e}^{-x}\cdots \left(iii\right)$
$\Rightarrow \frac{d^{3}y}{d{x}^3}=l\sin x-m\cos x+n{e}^{-x}\cdots \left(iv\right)$
Adding equation $\left(i\right)$ and $\left(iii\right),$ we get
$\frac{d^{2}y}{d{x}^2}+y=-2n{e}^{-x}\cdots \left(v\right)$
Adding equation $\left(ii\right)$ and $\left(iv\right),$ we get
$\frac{d^{3}y}{d{x}^3}+\frac{dy}{dx}=2n{e}^{-x}\cdots \left(vi\right)$
Adding equation $\left(v\right)$ and $\left(vi\right),$ we get
$\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$