calculus
posted by John .
Please help find the solution to the initial value problem dy/dx=(1+2cos^2 x)/y with(y>0), and y=1 when x=0. Thanks.

y dy = (1 + 2 cos^2 x)dx
d (y^2/2)= (1 + 2 cos^2 x) dx
y^2/2 = x + 2(x/2+sin2x/4) + c
y^2 = 2 x + 2x +sin 2x) + c
y^2 = 4 x + sin 2x + c
1 = c
y^ 2= 1 + sin 2x + 4x
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