# Treating differential operator as algebraic entity

• Apr 30th 2010, 05:24 AM
chappar
Treating differential operator as algebraic entity
I was reading e:the story of a number by Eli Maor, where he treats differential operator as just like any algebraic entity.

For example if we have a differential equation like y’’ + 5y’ - 6y = 0. This can be treaed as (D^2 + 5D – 6)y = 0. So, either y = 0 (trivial solution) or (D^2 + 5D – 6) = 0. Factoring out above equation we get (D-1)(D+6)= 0 with solutions as D = 1 and D = -6. Since D does not have any meaning on its own, multiplying by y on both the sides we get Dy = y and Dy = -6y for which the solutions are Ae^x and Be^-6x. Combining these 2 solutions we get Ae^x + Be^-6x.

Now my doubt is this approach break when we have an equation like D^2y = 0. Which means y = 0 (again trivial) or D^2 = 0 which means D = 0. Now Dy = y*0 = 0.

That means y = C ( a constant). The actual answer should be Cx. I know that it is stupidity to treat D^2 = 0 as D = 0, it led me to doubt the entire process of treating differential equation as algebraic equation. Can someone throw light on this?
• Apr 30th 2010, 10:47 AM
CaptainBlack
Quote:

Originally Posted by chappar
I was reading e:the story of a number by Eli Maor, where he treats differential operator as just like any algebraic entity.

For example if we have a differential equation like y’’ + 5y’ - 6y = 0. This can be treaed as (D^2 + 5D – 6)y = 0. So, either y = 0 (trivial solution) or (D^2 + 5D – 6) = 0. Factoring out above equation we get (D-1)(D+6)= 0 with solutions as D = 1 and D = -6. Since D does not have any meaning on its own, multiplying by y on both the sides we get Dy = y and Dy = -6y for which the solutions are Ae^x and Be^-6x. Combining these 2 solutions we get Ae^x + Be^-6x.

Now my doubt is this approach break when we have an equation like D^2y = 0. Which means y = 0 (again trivial) or D^2 = 0 which means D = 0. Now Dy = y*0 = 0.

That means y = C ( a constant). The actual answer should be Cx. I know that it is stupidity to treat D^2 = 0 as D = 0, it led me to doubt the entire process of treating differential equation as algebraic equation. Can someone throw light on this?

$D$ is an operator which under some circumstances satisfies common algebra's rules because what they are doing is shorthand for something else. But $D^2=0$ does not imply $D=0$. This is in fact all bad practice unless you know what you are doing, which is why we teach people to use Laplace Transform methods for this sort of thing.

CB