- #1
BiGyElLoWhAt
Gold Member
- 1,622
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So I had a quiz today, and one of the questions was pretty easy, pretty straight forward.
Show that ##t^2e^{9t}## is a solution of ##(D-9)^3y##
Foil it out, plug in y, and you're done.
Well I tried doing something else, that (at least in my mind) should have worked, but it didn't.
I said "Oh, well I'll just rewrite y as ##(y^{1/3})^3##, distribute it through, and solve ##D[y^{1/3}] - 9y^{1/3} = 0##
It definitely does not. What's even more interesting is that, upon distributing the y^1/3, depending on whether you solve the above equation, or you foil out ##(D[y^{1/3}] - 9y^{1/3})^3## you get 2 different sets of solutions. The given function definitely works when you foil out your "auxiliary equation" with the differential operator, but you get 2 different non-zero functions depending on how you choose to evaluate it. I applied the chain rule, so that's not it. I would write it out, but it gets kind of messy. Is there a reason why I can't do this? We're treating this diff eq. as a polynomial, so why can't I do polynomial stuff to it?
Chain rule as such ##D[y^{1/3}] = \frac{1}{3}y^{-\frac{2}{3}}y'##
Since D is a linear operator, can't I do this? Obviously not, as it doesn't work... but why?
Show that ##t^2e^{9t}## is a solution of ##(D-9)^3y##
Foil it out, plug in y, and you're done.
Well I tried doing something else, that (at least in my mind) should have worked, but it didn't.
I said "Oh, well I'll just rewrite y as ##(y^{1/3})^3##, distribute it through, and solve ##D[y^{1/3}] - 9y^{1/3} = 0##
It definitely does not. What's even more interesting is that, upon distributing the y^1/3, depending on whether you solve the above equation, or you foil out ##(D[y^{1/3}] - 9y^{1/3})^3## you get 2 different sets of solutions. The given function definitely works when you foil out your "auxiliary equation" with the differential operator, but you get 2 different non-zero functions depending on how you choose to evaluate it. I applied the chain rule, so that's not it. I would write it out, but it gets kind of messy. Is there a reason why I can't do this? We're treating this diff eq. as a polynomial, so why can't I do polynomial stuff to it?
Chain rule as such ##D[y^{1/3}] = \frac{1}{3}y^{-\frac{2}{3}}y'##
Since D is a linear operator, can't I do this? Obviously not, as it doesn't work... but why?