Revisiting Differential Operators.

 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 )   =    (x^n+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)