Question

The solution of $\frac{d^{2}x}{d{y}^2}-x=k$, where $k$ is a non-zero constant, vanishes when $y=0$ and tends of finite limit as $y$ tends to infinity, is

• A. $x=k(1+e^{-y})$
• B. $x=k(e^y+e^{-y}-2)$
• C. $x=k(e^{-y}-1)$
• D. $x=k(e^y-1)$

Answer 1

Answer: (C)

$x=k(e^{-y}-1)$

We can write given differential equation as,
$(D^2-1)x=k...\left(i\right)$
where, $D=\frac{d}{dy}$
Its auxiliary equation is $m^2-1=0$, so that
$m=1,-1$
Hence, $CF=C_1{e}^y+C_2{e}^{-y}$
where $C_1,C_2$ are arbitrary constants
Now, also $PI=\frac{1}{D^2-1}k$
$=k.\frac{1}{D^2-1}{e}^{0.y}$
$=k.\frac{1}{0^2-1}{e}^{0.y}=-k$
So, solution of eq. (i) is
$x=C_1{e}^y+C_2{e}^{-y}-k...\left(ii\right)$
Given that $x=0$, when $y=0$
So, $0=C_1+C_2-k$ (From (ii))
$\Rightarrow C_1+C_2=k....\left(iii\right)$
Multiplying both sides of eq. (ii) by $e^{-y}$, we get
$x.e^{-y}=C_1+C_2{e}^{-2y}-k{e}^{-y}...\left(iv\right)$
Given that $x\to m$ when $y\to \infty ,m$ being a finite quantity.
So, eq (iv) becomes
$x\times 0=C_1+C_2\times 0-(k\times 0)$
$\Rightarrow C_1=0....\left(v\right)$
From eqs. (iv) and (v), we get
$C_1=0$ and $C_2=k$
Hence, eq. (ii) becomes
$x=k{e}^{-y}-k=k(e^{-y}-1)$ which is the required solution.