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

We can write given differential equation as, $\left(\mathrm{D}^{2}-1\right) \mathrm{x}=\mathrm{k}$ ...(i) where, $D \equiv \frac{d}{d y}$ Its auxiliary equation is $m^{2}-1=0$, so that $\mathrm{m}=1,-1$ Hence, $C F=C_{1} e^{y}+C_{2} e^{-y}$. where $\mathrm{C}_{1}, \mathrm{C}_{2}$ are arbitrary constants Now, also $\mathrm{PI}=\frac{1}{\mathrm{D}^{2}-1} \mathrm{k}$ $\begin{array}{l} =\mathrm{k} \cdot \frac{1}{\mathrm{D}^{2}-1} \mathrm{e}^{0 . \mathrm{y}} \\ =K \cdot \frac{1}{0^{2}-1} e^{0 . \mathrm{y}}=-K \end{array}$ So, solution of eq. (i) is $x=C_{1} e^{y}+C_{2} e^{-y}-k$ ...(ii) Given that $x=0$, when $y=0$ So, $0=\mathrm{C}_{1}+\mathrm{C}_{2}-\mathrm{k} \quad$ (From (ii)) $\Rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}=\mathrm{k}$ ...(iii) Multiplying both sides of eq. (ii) by $\mathrm{e}^{-\mathrm{y}}$, we get x. $\mathrm{e}^{-\mathrm{y}}=\mathrm{C}_{1}+\mathrm{C}_{2} \mathrm{e}^{-2 \mathrm{y}}-\mathrm{ke}^{-\mathrm{y}}$ ...(iv) Given that $x \rightarrow m$ when $y \rightarrow \infty, m$ being a finite quantity. So, eq (iv) becomes $\mathrm{x} \times 0=\mathrm{C}_{1}+\mathrm{C}_{2} \times 0-(\mathrm{k} \times 0)$ $\Rightarrow \quad \mathrm{C}_{1}=0$ ...(v) From eqs. (iv) and (v), we get $\mathrm{C}_{1}=0$ and $\mathrm{C}_{2}=\mathrm{k}$ Hence, eq. (ii) becomes $\mathrm{x}=\mathrm{ke}^{-\mathrm{y}}-\mathrm{k}=\mathrm{k}\left(\mathrm{e}^{-\mathrm{y}}-1\right)$