Differential Operators are basically shorthand ways of writing derivatives and by extension writing more concise differential equations. In general:
(D x y) n = dn y / dxn n= 1, 2, 3….
Then: (D x y) 2 = d2 y/ dx2
(D x y) 3 = d3 y/ dx3
And so forth.
For example: d4y/ dt4
Can be rewritten: (D t y) 4
And: 2d3y/ dt3
Can be rewritten: 2(D t y)3
Then rewrite the differential equation:
d4y/ dt4 – 2d3y/ dt3 –7 d2y/ dt2 + 20 dy/dt –12 y = 0
Using differential operators.
Using the preceding hints for the notation, we have:
(D t y) 4 - 2(D t y)3 - 7 (D t y) 2 + 20 (D t y) - 12y = 0
Of special interest is the inverse differential operator, viz.
(D-1 x n ) = ( xn+1 ) / (n + 1)
Similarly for higher order::
(D-2 x n ) = (D-1 ) ( xn+1 ) / (n + 1) =
xn+2 / (n +1 ) (n + 2)
And for higher order (k) in general:
(D-k x n ) = (xn+k ) / (n+ 1) (n + 2)...(n + k)
Thus, 1/D stands for the integral, e.g. ∫ F(x)dx but with denominator corrected for as given by inverse formulae above.
And 1/ D n stands for successive (n) integration.
Suggested Problems:
1) Rewrite the DE below with differential operators:
5 d5y/
dt5 - 10 dy/dt –25y = 0
2) Rewrite the DE below in standard derivative form and solve:
(D x
y) = 3x2 - 1
3) Does (D x y) 2 = D 2 x y ? Explain.
4) Evaluate each of the following:
i) (D x y) (ln e x / 1+ e x)
ii) (D x y) ½ ( e x - e -x)
iii) D-1 ( 11 x2 )
iv) (D x y) ( e x ln x)
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