Question

Reduce the following differential equation to first-order and solve it. ​

Answer 1

Given that y₁(x) = x is a solution to the given second-order ODE, assume a second solution of the form

y₂(x) = y₁(x) v(x) = x v(x)

Take the first two derivatives and substitute this solution into the ODE:

y₂ = x v

y₂' = x v' + v

y₂'' = (x v'' + v' ) + v' = x v'' + 2 v'

→   (1 - x ²) y₂'' - 2 x y₂' + 2 y₂ = 0

→   (1 - x ²) (x v'' + 2 v' ) - 2 x (x v' + v) + 2 x v = 0

→   (x - x ³) v'' + (2 - 4 x ²) v' = 0

Next, reduce the order by substituting w(x) = v'(x) and w'(x) = v''(x) :

→   (x - x ³) w' + (2 - 4 x ²) w = 0

and now you have a linear equation in w. Solve for w(x) however you like; I'll just separate variables.

Write w'(x) = dw/dx, then separate variables as

(x - x ³) dw/dx + (2 - 4 x ²) w = 0

(x - x ³) dw/dx = (4x ² - 2) w

dw/w = (4x ² - 2) / (x - x ³) dx

Integrate both sides:

∫ dw/w = (4x ² - 2) / (x - x ³) dx

ln|w| = ∫ (4x ² - 2) / (x - x ³) dx

For the remaining integral, expand the integrand into partial fractions.

x - x ³ = x (1 - x ²) = x (1 - x) (1 + x)

→   (4x ² - 2) / (x - x ³) = a /x + b / (1 - x) + c / (1 + x)

→   4x ² - 2 = a (1 - x ²) + b x (1 + x) + c x (1 - x)

→   4x ² - 2 = (-a + b - c) x ² + (b + c) x + a

→   -a + b - c = 4, b + c = 0, a = -2

→   a = -2, b = 1, c = -1

→   (4x ² - 2) / (x - x ³) = -2/x + 1 / (1 - x) - 1 / (1 + x)

Then

ln|w| = ∫ (-2/x + 1 / (1 - x) - 1 / (1 + x)) dx

ln|w| = -2 ln|x| - ln|1 - x| - ln|1 + x| + C

ln|w| = -ln|x ² (1 - x ²)| + C

exp(ln|w|) = exp(-ln|x ² (1 - x ²)| + C )

w = C / (x ² (1 - x ²))

v' = C / (x ² (1 - x ²))

Solve for v(x) by integrating both sides, expanding the right side into partial fractions again:

x ² (1 - x ²) = x ² (1 - x) (1 + x)

→   1 / (x ² (1 - x ²)) = a / x + b / x ² + c / (1 - x) + d / (1 + x)

→   1 = a x (1 - x ²) + b (1 - x ²) + c x ² (1 + x) + d x ² (1 - x)

→   1 = (-a + c - d) x ³ + (-b + c + d) x ² + a x + b

→   -a + c - d = 0, -b + c + d = 0, a = 0, b = 1

→   a = 0, b = 1, c = 1/2, d = 1/2

→   1 / (x ² (1 - x ²)) = 1/x ² + 1/2 (1 / (1 - x) + 1 / (1 + x))

dv/dx = C (1/x ² + 1/2 (1 / (1 - x) + 1 / (1 + x)))

dv = C (1/x ² + 1/2 (1 / (1 - x) + 1 / (1 + x))) dx

Integrate both sides:

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

v = C₁ (-1/x - 1/2 ln|1 - x| + 1/2 ln|1 + x|) + C₂

v = C₁ (1/2 ln|(1 + x) / (1 - x)| - 1/x) + C₂

Solve for y₂(x) by multiplying both sides by x :

y₂ = C₁ (1/2 x ln|(1 + x) / (1 - x)| - 1) + C₂ x

Now, y₁(x) = x, so we can omit the last term from the second solution, giving us

y₂(x) = C (1/2 x ln|(1 + x) / (1 - x)| - 1)