please help me find the solution of this initial value problem?

Question

dy/dx = cos(4x)/3-sin(4x) when y=6 and x=0 i am a bit stuck any help welcomed

Answer 1

dy/dx=cos(4x)/3 -sin(4x)
y=sin(4x)/12+cos(4x)/4+C
6=0+1/4+C
C=5 3/4
y=sin(4x)/12+cos(4x)/4)+5 3/4

Answer 2

The integral of cos(4x)/3 dx is sin(4x)/12 + c
The integral of -sin(4x)dx is cos(4x)/4 + c

Now let's add them and get y=sin(4x)/12 + cos(4x)/4 + c

put x=0 and get
6 = sin (4 * 0)/12 + cos( 4 * 0)/4 + c = 0 + 1/4

c = 5.75

Answer 3

well... i'm gonna attempt this purely because i'm bored lol and i'm off school ill so here goes...
d6/d0=cos4/3-sin4 ...

cos4=0.99756405

sin4=0.069756473 so 3-0.069756473=2.930243527

d6/d0=0.99756405/2.930243527
so, 0.99756405/2.930243527 = 0.34043725

d6/d0= 0.34043725

then you have gota find out what d is :S so maybe you divide the 0.34043725 by 6 to get the value of d...

0.34043725/6 = 0.056739541 so d=0.056739541
now replace the d's in the equation with 0.056739541 to see if it works...

6x0.056739541=0.34043725/0x0.34043725 <(which equals zero)
=0.34043725

soooo... now it is equal on both sides of the equals sign ecause the sum of d6/d0 is 0.34043725, and the sum of cos(x4)/3-sin(x4) is also 0.34043725.

the final equation is 0.34043725/0.34043725

and the value of d is 0.056739541

i hope this helps, its probably wrong but i tried!! i am only in yr11 so this is pretty hard stuff for me so don't trust my answer to be completely correct, cya, Jess x x