# A function n(x) satisfies the differential equation $\dfrac{d^2n(x)}{dx^2} - \dfrac{n(x)}{L^2} = 0$ where L is a constant. The boundary conditions a

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A function n(x) satisfies the differential equation $\dfrac{d^2n(x)}{dx^2} - \dfrac{n(x)}{L^2} = 0$ where L is a constant. The boundary conditions are: n(0)=K and n(∞) = 0. The solution to this equation is
1. n(x) = K exp(x/L)
2. n(x) = K exp(-x/√L)
3. n(x) = K2 exp(-x/L)
4. n(x) = K exp(-x/L)

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Correct Answer - Option 4 : n(x) = K exp(-x/L)

Concept:

The general solution of different types of roots for a particular DE is shown below:

 Roots of AE Nature General solution a, b, c ⋯ Real & distinct y(x) = C1eax + C2ebx + ⋯ a, a, a, a ⋯ Real & repeated y(x) = C1eax + C2xeax + C3 x2eax +⋯ α ± iβ Complex y(x) = eαx[C1 cos βx + C2 sin βx] α ± iβ n times Complex & repeated y(x) = eαx[(C1+C2x + ⋯ Cnxn-1) cos βx + D1+D2x + ⋯ Dnxn-1) sin βx ] ± iβ purely imaginary y(x) = C1 cos βx + C2 sin βx irrational α ± √β Irrational y(x) = eαx[C1 cosh √βx + C2 sinh √βx]

Calculation:

Given DE is:

$\frac{{{d^2}n\left( x \right)}}{{d{x^2}}} - \frac{{n\left( x \right)}}{{{L^2}}} = 0$         ---(1)

n(0) = k       ---(2)

n(∞) = 0      ---(3)

From equation (1) we can write the DE as:

$\left( {{D^2} - \frac{1}{{{L^2}}}} \right)n = 0$

The auxiliary equation is:

${D^2} - \frac{1}{{{L^2}}} = 0$

D = ± 1/L

Roots are real & distinct

The general solution is:

n(x) = c1ex/L + c2e-x/L        ---(4)

$\frac{{dn\left( x \right)}}{{dx}} = \frac{1}{L}{c_1}{e^{\frac{x}{L}}} - \frac{{{c_2}}}{L}{e^{ - \frac{x}{L}}}$      ---(5)

Using equations (2),  (4) becomes

K = c1 + c2     ---(6)

Using (3), (4) becomes

0 = c1(∞) + c2(0)

c1(∞) = 0

c1 = 0

From (6)

c2 = k

The solution of the equation (1) is:

n(x) = ke-x/L