Question

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)

Answer 1

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) = c₁e^x/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