GATE1998(2): Higher Order Differential Equation

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Question: GATE1998 (2 Marks)

The radial displacement in a rotating disc is governed by the differential equation \(\displaystyle \frac{d^2u}{dx^2} + \frac{1}{x} \frac{du}{dx} – \frac{u}{x^2} = 8x \) where u is the displacement and x is the radius. If u = 0 at x = 0, and u = 2 at x = 1, calculate the displacement at x = \(\frac{1}{2} \)

Solution:

Multiplying throughout by x2

x2u’’ + xu’ + u = 8x3

This is a second order Euler-Cauchy Differential Equation

Take y = xt

y’ = txt-1

y’’ = t(t-1)xt-2

The differential equation becomes xt (t2 – 1) = 0

Values of t = +1 and -1

Solution: u = C1x + C2/x

Applying boundary conditions:

u(0) = 0

C2 = 0

u(1) = 2

2 = C1 x 1

C1 = 2

The complementary function becomes u = 2x

Particular Integral \(\displaystyle P.I = \frac{8x^3}{D^2-1} \)

\(\displaystyle \frac{8x^3}{D^2-1} \) = – (1 – D2)-1[8x3]

= – (1 + D2 + D4) [8x3]

= –8x3 – 48x

Complete solution is u = 2x – 8x3 – 48x = –8x3 – 46x

u(1/2) = –1 – 23 = –24

Answer: The displacement at x = 1/2 is – 24