Question

Find the solution to the initial value problem \[f '(x)=4(\cos u-\sin2 u); f(\frac{ \pi }{6 })=0\]

Answer 1

So far I have \[f(x)=\sin 4u+2\cos 2u+C\]but when I try to plug in pi/6 for u I'm not getting the correct answer for C which should be 3 (according to the back of the book)

Answer 2

-3 that is

Answer 3

@ZeHanz

Answer 4

The factor 4 can be ignored. Put it as first factor in f(x):

\(f(x)=4( \sin u +\frac{1}{2}\cos2u)\). Oh, now I can see your error:

\(f(x)=4\sin x + 2\cos x +C\). You typed sin4x...

Answer 5

This function will give you C=-3, if you set \(f(\frac{\pi}{6})=0\).

Answer 6

oh you're right

Answer 7

LOL thanks alot

Answer 8

YW!
I make these errors myself all the time!

Answer 9

Good thing you caught this one. I was really confused