For the differential equation,the general sol | vedclass

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(B) Given equation: $x \cos \left( \frac{y}{x} \right) (y dx + x dy) = y \sin \left( \frac{y}{x} \right) (x dy - y dx)$ Divide both sides by $xy \cos

Vedclass · Accepted Answer

$xy \cos \left( \frac{y}{x} \right) = c$

Vedclass · Answer

(B) Given equation: $x \cos \left( \frac{y}{x} ight) (y dx + x dy) = y \sin \left( \frac{y}{x} ight) (x dy - y dx)$ Divide both sides by $xy \cos \left( \frac{y}{x} ight)$: $\frac{y dx + x dy}{y} = 	an \left( \frac{y}{x} ight) \frac{x dy - y dx}{x} $ $\frac{y dx + x dy}{xy} = 	an \left( \frac{y}{x} ight) \left( \frac{x dy - y dx}{x^2} ight) $ Note that $d(xy) = y dx + x dy$ and $d\left( \frac{y}{x} ight) = \frac{x dy - y dx}{x^2}$. So,$\frac{d(xy)}{xy} = 	an \left( \frac{y}{x} ight) d\left( \frac{y}{x} ight) $ Integrating both sides: $\int \frac{d(xy)}{xy} = \int 	an \left( \frac{y}{x} ight) d\left( \frac{y}{x} ight) $ $\ln |xy| = \ln \left| \sec \left( \frac{y}{x} ight) ight| + \ln |c| $ $xy = c \sec \left( \frac{y}{x} ight) $ Thus,$x = \frac{c}{y} \sec \left( \frac{y}{x} ight)$ is not the form,but $xy = c \sec \left( \frac{y}{x} ight)$ is equivalent to $xy \cos \left( \frac{y}{x} ight) = c$.

For the differential equation,the general solution for $x \cos \left( \frac{y}{x} \right) (y dx + x dy) = y \sin \left( \frac{y}{x} \right) (x dy - y dx)$,(where $c$ is the constant of integration) is

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