# maths

posted by .

Can any one help with the following please:
Solve the initial-value problem:
problem:
dy/dx: 4(3cos(6x)-1)/1+2x-sin(6x) y=12 when x=0

• Calc-IVP -

Do 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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