A curve passes through the point left(1, frac | vedclass

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(A) The slope of the curve is given by $\frac{dy}{dx} = \frac{y}{x} + \sec\left(\frac{y}{x}\right)$.This is a homogeneous differential equation. Let $

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$\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}$

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(A) The slope of the curve is given by $\frac{dy}{dx} = \frac{y}{x} + \sec\left(\frac{y}{x}\right)$.This is a homogeneous differential equation. Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = v + \sec(v)$.This simplifies to $x\frac{dv}{dx} = \sec(v)$,which can be rewritten as $\cos(v) dv = \frac{1}{x} dx$.Integrating both sides: $\int \cos(v) dv = \int \frac{1}{x} dx$.This gives $\sin(v) = \ln|x| + C$.Substituting $v = \frac{y}{x}$ back,we get $\sin\left(\frac{y}{x}\right) = \ln|x| + C$.The curve passes through $\left(1, \frac{\pi}{6}\right)$,so $\sin\left(\frac{\pi/6}{1}\right) = \ln(1) + C$.Since $\sin(\pi/6) = 1/2$ and $\ln(1) = 0$,we find $C = 1/2$.Thus,the equation of the curve is $\sin\left(\frac{y}{x}\right) = \ln x + \frac{1}{2}$.

$A$ curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right)$,$x>0$. Then the equation of the curve is

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