The solution of x2dydx−xy=1+cos(yx) is tan(ypx)=C−1qx2, when C is arbitrary constant then the value of ∫0p+q{x}dx is ______({.}is fractional part) s

x(xdy−ydx)=(1+cosyx)dx d(yx)1+cos(yx)=dyx312∫sec2(y2x)d(yx)=∫dyx312tan(y2x)12=−12x2+Ctan(y2x)=C−12x2∫04{x}dx=42=2

The solution of x2dydx−xy=1+cos(yx) is tan(ypx)=C−1qx2, when C is arbitrary constant then the value of ∫0p+q{x}dx is ______({.}is fractional part) s

x(xdy−ydx)=(1+cosyx)dx d(yx)1+cos(yx)=dyx312∫sec2(y2x)d(yx)=∫dyx312tan(y2x)12=−12x2+Ctan(y2x)=C−12x2∫04{x}dx=42=2

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The solution of $x^{2} \frac{d y}{d x} - x y = 1 + \cos (\frac{y}{x})$ is $\tan (\frac{y}{p x}) = C - \frac{1}{q x^{2}},$ when C is arbitrary constant then the value of $\int_{0}^{p + q} {x} d x i s______({.} i s f r a c t i o n a l p a r t)$ s

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Detailed Solution

$x (x d y - y d x) = (1 + \cos \frac{y}{x}) d x \begin{array}{l} \frac{d (\frac{y}{x})}{1 + \cos (\frac{y}{x})} = \frac{d y}{x^{3}} \\ \frac{1}{2} \int \sec^{2} (\frac{y}{2 x}) d (\frac{y}{x}) = \int \frac{d y}{x^{3}} \\ \frac{1}{2} \frac{\tan (\frac{y}{2 x})}{\frac{1}{2}} = \frac{- 1}{2 x^{2}} + C \\ \tan (\frac{y}{2 x}) = C - \frac{1}{2 x^{2}} \\ \int_{0}^{4} {x} d x = \frac{4}{2} = 2 \end{array}$