Shazz

New member
y=e^-x (cosx+sinx). Prove that Y4+4y=0

Subhotosh Khan

Super Moderator
Staff member
y=e^-x (cosx+sinx). Prove that Y4+4y=0

You have both "y" and "Y" in the statement. Are those different?

You wrote "Y4".

Did you mean Y4 or Y*4 or Y4 as a variable name.

Shazz

New member

You have both "y" and "Y" in the statement. Are those different?

You wrote "Y4".

Did you mean Y4 or Y*4 or Y4 as a variable name.
It's y4 meant 4th derivative

Subhotosh Khan

Super Moderator
Staff member
It's y4 meant 4th derivative
You do not have y4

You wrote Y4.

topsquark

Senior Member
y=e^-x (cosx+sinx). Prove that Y4+4y=0
The usual way to write a derivative is [imath]y^{(4)}[/imath].

What is the problem, then? Is it taking the derivatives? Use the product rule.
[imath]y' = -e^{-x} ( cos(x) + sin(x) ) + e^{-x} (-sin(x) + cos(x) )[/imath]

HInt: The next derivative is similar, just with a couple of signs and sines and cosines switched. You'll end up adding a couple of terms and get 2's. The 3rd and 4th derivatives do the same thing. You won't have any more than 2 terms in your (simplified) derivative so it's not as bad as it might look at first.

Can you continue?

-Dan