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solve the initial value problem dy/dx = sin(4x)/3 + cos (4x) where y=5 when x=0
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It is given that `dy/dx = sin(4x)/3 + cos(4x)` and y = 5 when x = 0.
`dy/dx = sin(4x)/3 + cos(4x)`
=> `dy = (1/3)*sin 4x dx + cos 4x dx`
`int dy = (1/3)*int sin 4x dx + int cos 4x dx`
=> `y = (1/3)*(-1/4)*(cos 4x) + (1/4)*sin 4x + C`
As y = 5 when x = 0
`5 = (sin 0)/4 - (cos 0)/12 + C`
=> `5 + 1/12 = C`
=> `C = 61/12`
This gives `y = (sin 4x)/4 - (cos 4x)/12 + 61/12`
Posted by justaguide on February 28, 2012 at 9:10 PM (Answer #1)
The 3 + Cos(4x) should all be on the bottom of the equation, so does this mean that I can do a third of sin(4x)/cos(4x) and still get the correct answer?
Posted by debzb1 on February 28, 2012 at 10:16 PM (Answer #2)
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