- #1

Exulus

- 50

- 0

[itex]\frac{d^2y}{dx^2} + 4y = \sin{2x}[/itex]

Everything I seem to try for y ends up canceling itself out, as the second diffferential is always the negative of what you begin with, and the 2x part of the sin means you get a factor of 4 in the second differential, which then cancels with the 4y. Eg:

Trying [itex]y = c \sin{2x}[/itex]

[itex]\frac{dy}{dx} = 2c \cos{2x}[/itex]

[itex]\frac{d^2y}{dx^2} = -4c \sin{2x}[/itex]

Substituting these into the LHS of my question gives:

[itex] -4c \sin{2x} + 4c \sin{2x} = 0[/itex]

The only thing i can think of now is that its a complex solution..but we haven't been taught that yet!

Any help appreciated :(

(ps no idea why the itex stuff is coming out so small..sorry about that).