Question

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

A
x=k(1+ey)
B
x=k(ey+ey2)
C
x=k(ey1)
D
x=k(ey1)

Solution

## The correct option is B $$x = k(e^{-y} - 1)$$We can write given differential equation as,$$(D^{2} - 1)x = k ... (i)$$where, $$D = \dfrac {d}{dy}$$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 constantsNow, also $$PI = \dfrac {1}{D^{2} - 1}k$$$$= k.\dfrac {1}{D^{2} - 1} e^{0.y}$$$$= k.\dfrac {1}{0^{2} - 1}e^{0.y} = -k$$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, $$0 = C_{1} + C_{2} - k$$ (From (ii))$$\Rightarrow C_{1} + C_{2} = k .... (iii)$$Multiplying both sides of eq. (ii) by $$e^{-y}$$, we get$$x.e^{-y} = C_{1} + C_{2}e^{-2y} - ke^{-y} ... (iv)$$Given that $$x\rightarrow m$$ when $$y\rightarrow \infty, m$$ being a finite quantity.So, eq (iv) becomes$$x\times 0 = C_{1} + C_{2} \times 0 - (k\times 0)$$$$\Rightarrow C_{1} = 0 ....(v)$$From eqs. (iv) and (v), we get$$C_{1} = 0$$ and $$C_{2} = k$$Hence, eq. (ii) becomes$$x = ke^{-y} - k = k(e^{-y} - 1)$$ which is the required solution.Mathematics

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