# u and v are solutions

• Apr 6th 2010, 10:18 PM
onemore
u and v are solutions
Clearly u = cos x and v = sin x are solutions of y'' + y = 0. Show that $z = u^2$ and z = uv are solutions of z''' + 4z' = 0

I dont understand how to start this off. Im not asking for the answer just trying to understand what is going on or what to do.
• Apr 7th 2010, 02:09 AM
tom@ballooncalculus
Is what's meant to be clear, clear? Just in case a picture helps...

http://www.ballooncalculus.org/asy/diffChain/f2.png

... differentiating downwards.

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• Apr 7th 2010, 03:57 AM
HallsofIvy
Quote:

Originally Posted by onemore
Clearly u = cos x and v = sin x are solutions of y'' + y = 0. Show that $z = u^2$ and z = uv are solutions of z''' + 4z' = 0

I dont understand how to start this off. Im not asking for the answer just trying to understand what is going on or what to do.

$z= u^2= cos^2(x)$. What is $(cos^2(x))'''+ 4(cos^2(x))'$?

$z= uv= cos(x)sin(x)$. What is $(cos(x)sin(x))'''+ 4(cos(x)sin(x))'$?