Every curve represented by the general soluti | vedclass

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(C) Let the slope of the first curve be $m_1 = \frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$.Let the slope of the second curve be $m_2 = \frac{dy}{d

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$\frac{3\pi}{2}$

Vedclass · Answer

(C) Let the slope of the first curve be $m_1 = \frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$.Let the slope of the second curve be $m_2 = \frac{dy}{dx} = -\frac{y^3 e^{y^2-5}}{x \log x}$.Now,calculate the product of the slopes: $m_1 	imes m_2 = \left(\frac{x \log x}{y^3 e^{y^2-5}}ight) 	imes \left(-\frac{y^3 e^{y^2-5}}{x \log x}ight) = -1$.Since the product of the slopes is $-1$,the curves are orthogonal,meaning the angle of intersection is $	heta = \frac{\pi}{2}$.Finally,calculate $4	heta - \frac{\pi}{2} = 4\left(\frac{\pi}{2}ight) - \frac{\pi}{2} = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.

Every curve represented by the general solution of $\frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$ cuts every curve represented by the general solution of $\frac{dy}{dx} + \frac{y^3 e^{y^2-5}}{x \log x} = 0$ at an angle $\theta$. Then,$4\theta - \frac{\pi}{2} =$

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