Existence and Uniqueness of Solutions of Second Order D.E.'s

In a Nut Shell: Under restrictions solutions exist for linear second order differential

equations that are unique throughout an interval. The theorem below details these

restrictions.

Theorem for a general second order, ordinary differential equation of the form:

y'' + p(x) y' + q(x) y = g(x)

with initial conditions y(x_o) = y_o, y'(x_o) = y'_o

where p(x), q(x), and g(x) are continuous functions on an open interval I that contains

the point x_o .

Then there is only one solution , y, that exists throughout the interval I. i.e.

A solution exists for the differential equation .

The differential equation has only one solution. The solution is unique.

The solution exists throughout the interval I where the coefficients are continuous

and twice differentiable in the interval.

Example: Determine the longest interval, I, for the following initial value application.

(x ˗ 5) y'' + x y' + ( ln x ) y = 0 y(3) = 0, y'(3) = 2

y'' + (x/(x ˗ 5) y' + lnx / (x ˗ 5 ) y = 0

x/(x ˗ 5) continuous except for x = 5

ln x / (x ˗ 5) continuous except for x = 5 with restriction x > 0

So the longest interval containing x = 3 is ( 0 , 5 ) (result)

Example: Determine the longest interval, I, for the following differential equation.

( x² ˗ 36) y ⁽⁴⁾ + 6 x² y ''' + 18 y = 0

y ⁽⁴⁾ + [ 6 x² / ( x² ˗ 36) ] y ''' + [ 18 / ( x² ˗ 36) ] y = 0

y ⁽⁴⁾ + [ 6 x² / ( x ˗ 6) (x + 6)] y ''' + [ 18 / ( x ˗ 6) (x+6) ] y = 0

Points of discontinuity are at x = 6 and at x = ˗ 6

So the longest interval contains ( ˗ ∞ , ˗ 6 ) ( ˗6 , 6 ) (6 , ∞ ) (result)

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