### 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 x^{2}

x^{2}u’’ + xu’ + u = 8x^{3}

This is a second order Euler-Cauchy Differential Equation

Take y = x^{t}

y’ = tx^{t-1}

y’’ = t(t-1)x^{t-2}

The differential equation becomes x^{t }(t^{2} – 1) = 0

Values of t = +1 and -1

Solution: u = C_{1}x + C_{2}/x

Applying boundary conditions:

u(0) = 0

C_{2} = 0

u(1) = 2

2 = C_{1} x 1

C_{1} = 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 – D^{2})^{-1}[8x^{3}]

= – (1 + D^{2} + D^{4}) [8x^{3}]

= –8x^{3} – 48x

Complete solution is u = 2x – 8x^{3} – 48x = –8x^{3} – 46x

u(1/2) = –1 – 23 = –24