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

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$.This is a homogeneous differential equation. Let $v

Vedclass · Accepted Answer

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

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$.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$$x \frac{dv}{dx} = \sec v$$\cos v \, dv = \frac{1}{x} \, dx$.Integrating both sides:$\int \cos v \, dv = \int \frac{1}{x} \, dx$$\sin v = \log x + C$.Substituting $v = \frac{y}{x}$ back:$\sin \left(\frac{y}{x}\right) = \log x + C$.The curve passes through $\left(1, \frac{\pi}{6}\right)$,so:$\sin \left(\frac{\pi/6}{1}\right) = \log(1) + C$$\sin \left(\frac{\pi}{6}\right) = 0 + C \Rightarrow 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 given by $\frac{dy}{dx} = \frac{y}{x} + \sec \left(\frac{y}{x}\right)$,where $x > 0$. Then the equation of the curve is:

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