# calculus

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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.

• calculus -

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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