Question

The order of the differential equation whose general solution is given by
$y=(C_1+C_2)\cos (x+C_3)-C_4{e}^{x+C_5}$
where, $C_1,C_2,C_3,C_4,C_5$ are arbitrary constants, is

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

Answer: (C)

$3$

We have,
$y=(C_1+C_2)\cos (x+C_3)-C_4{e}^{x+C_5}$ ..... (i)
$\Rightarrow y=(C_1+C_2)\cos (x+C_3)-C_4{e}^x\cdot e^{C_5}$
Now, let $C_1+C_2=A,C_3=B,C_4{e}^{C_5}=C$
$\Rightarrow y=A\cos (x+B)-C{e}^x$ .... (ii)
$\Rightarrow \text{three arbitrary constant}$
$\therefore \text{degree 3}$
On differentiating (ii) w.r.t. $x$, we get
$\frac{dy}{dx}=-A\sin (x+B)-C{e}^x$ .... (iii)
Again, differentiating w.r.t. $x$, we get
$\frac{d^{2}y}{d{x}^2}=-A\cos (x+B)-C{e}^x$ .... (iv)
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-y-2C{e}^x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}+y=-2C{e}^x$ .... (v)
Again, differentiating w.r.t. $x$, we get
$\frac{d^{3}y}{d{x}^3}+\frac{dy}{dx}=-2C{e}^x$ .... (vi)
$\Rightarrow \frac{d^{3}y}{d{x}^3}+\frac{dy}{dx}=\frac{d^{2}y}{d{x}^2}+y$ [from Eq. (v)]
which is a differential equation of order $3$.