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December 19, 2014

December 19, 2014

Posted by **Claire** on Tuesday, February 8, 2011 at 4:40am.

Solve the initial-value problem:

problem:

dy/dx: 4(3cos(6x)-1)/1+2x-sin(6x) y=12 when x=0

- Calc-IVP -
**MathMate**, Tuesday, February 8, 2011 at 9:02amDo you mean

dy/dx= 4(3cos(6x)-1) / (1+2x-sin(6x)) ?

Substitute

u=(1+2x-sin(6x))

du=(-6cos(6x)+2)

so

dy/dx = -2∫du/u

y=-2ln(1+2x-sin(6x))+C

Initial condition: y(0)=12

y(0)=12=-2ln(1+2(0)-sin(6(0)))+C

12=-2ln(1)+C

12=0+C

=> C=12

So

y=-2ln(1+2x-sin(6x))+12

Check the result by differentiation.

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