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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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debzb1 | (Level 1) eNoter

Posted February 28, 2012 at 7:37 PM via web

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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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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 28, 2012 at 9:10 PM (Answer #1)

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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`

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debzb1 | (Level 1) eNoter

Posted February 28, 2012 at 10:16 PM (Answer #2)

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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?

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