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Q.
The solution of $ \frac{d^{2}x}{dy^{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
We can write given differential equation as,
$ \left(D^{2} - 1\right) 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. $\left(i\right)$ 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 $\left(ii\right)$)
$\Rightarrow C_{1} + C_{2} = k....\left(iii\right)$
Multiplying both sides of eq. $\left(ii\right)$ by $e ^{-y}$we get
$x. e^{-y} = C_{1} + C_{2}e^{-2y} -ke^{-y} ....\left(iv\right)$
Given that $x \to m$ when $y \to \infty,$ m being a finite quantity.
So, eq $\left(iv\right)$ becomes
$x × 0 = C_{1} + C_{2} × 0 - \left(k × 0\right)$
$\Rightarrow C_{1} = 0....\left(v\right)$
From eqs. $\left(iv\right)$ and $\left(v\right)$, we get
$C_{1} = 0$ and $C_{2} = k$
Hence, eq. $\left(ii\right)$ becomes
$x = ke^{-y} - k = k\left(e^{-y} -1\right)$