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

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(A) The given differential equation is $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$.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 given differential equation is $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$.This is a homogeneous differential equation. Let $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.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 written as $\cos(v) dv = \frac{1}{x} dx$.Integrating both sides: $\int \cos(v) dv = \int \frac{1}{x} dx$,which gives $\sin(v) = \log(x) + c$.Substituting $v = \frac{y}{x}$ back,we get $\sin \left(\frac{y}{x}\right) = \log(x) + c$.The curve passes through $\left(1, \frac{\pi}{6}\right)$,so $\sin \left(\frac{\pi/6}{1}\right) = \log(1) + c$.Since $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\log(1) = 0$,we have $c = \frac{1}{2}$.Thus,the equation of the curve is $\sin \left(\frac{y}{x}\right) = \log(x) + \frac{1}{2}$.

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

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