maths
posted by Claire .
Can any one help with the following please:
Solve the initialvalue problem:
problem:
dy/dx: 4(3cos(6x)1)/1+2xsin(6x) y=12 when x=0

Do you mean
dy/dx= 4(3cos(6x)1) / (1+2xsin(6x)) ?
Substitute
u=(1+2xsin(6x))
du=(6cos(6x)+2)
so
dy/dx = 2∫du/u
y=2ln(1+2xsin(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+2xsin(6x))+12
Check the result by differentiation.