The solution of x2dydx−xy=1+cos ⁡yxis

x2dydx−xy=1+cos⁡yxor x(xdy−ydx)dx=1+cos⁡yxor xdy−ydxx21+cos ⁡yx=dxx3or ∫dyx1+cos ⁡yx=∫dxx3or 12∫dyxcos2 ⁡y2x=∫dxx3or 12∫sec2 ⁡y2x⋅dyx=∫dxx3or 12⋅tan ⁡y2x12=x−2−2+cor tan ⁡y2x+12x2=c

The solution of x2dydx−xy=1+cos ⁡yxis

x2dydx−xy=1+cos⁡yxor x(xdy−ydx)dx=1+cos⁡yxor xdy−ydxx21+cos ⁡yx=dxx3or ∫dyx1+cos ⁡yx=∫dxx3or 12∫dyxcos2 ⁡y2x=∫dxx3or 12∫sec2 ⁡y2x⋅dyx=∫dxx3or 12⋅tan ⁡y2x12=x−2−2+cor tan ⁡y2x+12x2=c

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The solution of $x^{2} \frac{dy}{dx} - xy = 1 + \cos \frac{y}{x}$ is

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$\tan (\frac{y}{2 x}) = c - \frac{1}{2 x^{2}}$

$\cos (\frac{y}{x}) = 1 + \frac{c}{x}$

$\tan \frac{y}{x} = c + \frac{1}{x}$

$x^{2} = (c + x^{2}) \tan \frac{y}{x}$

answer is A.

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Detailed Solution

$x^{2} \frac{dy}{dx} - xy = 1 + \cos \frac{y}{x}$
or $\frac{x (xdy - ydx)}{dx} = 1 + \cos \frac{y}{x}$
or $\frac{\frac{xdy - ydx}{x^{2}}}{1 + \cos \frac{y}{x}} = \frac{dx}{x^{3}}$
or $\int \frac{d (\frac{y}{x})}{1 + \cos \frac{y}{x}} = \int \frac{dx}{x^{3}}$
or $\frac{1}{2} \int \frac{d (\frac{y}{x})}{\cos^{2} \frac{y}{2 x}} = \int \frac{dx}{x^{3}}$
or $\frac{1}{2} \int \sec^{2} \frac{y}{2 x} \cdot d (\frac{y}{x}) = \int \frac{dx}{x^{3}}$
or $\frac{1}{2} \cdot \frac{\tan \frac{y}{2 x}}{\frac{1}{2}} = \frac{x^{- 2}}{- 2} + c$
or $\tan \frac{y}{2 x} + \frac{1}{2 x^{2}} = c$