Question

for given differential eqn find a particular solution
$x({ x}^{2} - 1)dy \div dx = 1$
when y = 0 and x =2

Answer 1

Solution:

The given differential equation is

x (x² - 1) dy/dx = 1

or, x (x + 1) (x - 1) dy/dx = 1

or, dy = dx/{x (x + 1) (x - 1)}

Integrating, we get

∫ dy = ∫ dx/{x (x + 1) (x - 1)} ... (1)

Let, 1/{x (x + 1) (x - 1)} = A/x + B/(x + 1) + C/(x - 1),

where A, B, C are constants

or, A (x + 1) (x - 1) + B x (x - 1) + C x (x + 1) = 1

Comparing among the coefficients, we get

A + B + C = 0 ... (2)

- B + C = 0

or, B = C ... (3)

- A = 1 or, A = - 1

From (2), we get

- 1 + B + C = 0

or, B + C = 1

or, B + B = 1, by (3)

or, 2B = 1

or, B = 1/2

Then from (3), we get C = 1/2

From (1), we can write

y = - ∫ dx/x + 1/2 ∫ dx/(x + 1) + 1/2 ∫ dx/(x - 1) + C,

where C is constant of integration

or, y = - logx + 1/2 log(x + 1) + 1/2 log(x - 1) + C

or, y = - logx+ 1/2 log{(x + 1) (x - 1)} + C

or, y = log[{√(x² - 1)} / x] + C

When y = 0 and x = 2, we get

0 = log[ {√(2² - 1)} / 2 ] + C

or, C = - log{ (√3)/2 }

Thus the required solution is

y = log[{√(x² - 1)} / x] - log{ (√3)/2 }

or, y = log[(2/√3){√(x² - 1)} / x]