Please help

Shazz said:
y=e^-x (cosx+sinx). Prove that Y4+4y=0
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Please review your statement and possibly correct it.

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.
 
Subhotosh Khan said:
Please review your statement and possibly correct it.

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.
Click to expand...
It's y4 meant 4th derivative
 
Shazz said:
y=e^-x (cosx+sinx). Prove that Y4+4y=0
Click to expand...
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
 
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