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A curve passes through the point (1, pi/6). At each point (x, y) with x > 0, the slope of the curve is given by y/x + sec(y/x). Find the equation of the curve.
- sin(y/x) = ln(x) + 1/2
- cosec(y/x) = ln(x) + 2
- sec(2y/x) = ln(x) + 2
- cos(2y/x) = ln(x) + 1/2
Correct answer: sin(y/x) = ln(x) + 1/2
Solution
dy/dx = y/x + sec(y/x). Let v=y/x: dy/dx = v + x*dv/dx. So v+x*dv/dx = v+sec(v) -> x*dv/dx = sec(v) -> cos(v)dv = dx/x. Integrating: sin(v) = ln(x)+C. At (1,pi/6): sin(pi/6)=ln(1)+C -> 1/2=0+C. So sin(y/x)=ln(x)+1/2.
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