The solution of xdy-x2-y2 - y-x2-y2-1 dx is-

The correct option is B y = x tan (c x) x/x^2+y^2 dy = ( #160;y/x^2+y^2 1 ) dx ⇒ #160;xdy ydx/x^2+y^2 = dx ⇒ tan^ 1 ...

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Basic Inverse Trigonometric Functions

The solution ...

Question

The solution of $\frac{x d y}{x^{2} + y^{2}} = (\frac{y}{x^{2} + y^{2}} - 1) d x$ is:

y = x c o t (c - x)

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c o s^{1} y / x = x + c

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y = x t a n (c - x)

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y^{2} / x^{2} = x t a n (c - x)

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\frac{x d y}{x^{2} + y^{2}} = (\frac{y}{x^{2} + y^{2}} - 1) d x

\frac{x d y}{x^{2} + y^{2}} = (\frac{y}{x^{2} + y^{2}} - 1) d x

x d x + y d y = \frac{x d y - y d x}{x^{2} + y^{2}}

2 {tan}^{- 1} \frac{y}{x} - 1 = x^{2} + y^{2} + c

d ({tan}^{- 1} \frac{y}{x}) = \frac{x d y - y d x}{x y}

\frac{x + \frac{d y}{d x}}{y - x \frac{d y}{d x}} = \frac{x {cos}^{2} (x^{2} + y^{2})}{y^{3}}

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

x d x + y d y = \frac{1}{2} d (x^{2} + y^{2})

\frac{y d x - x d y}{y^{2}} = d (\frac{x}{y})

sin 21^{\circ} = \frac{x}{y}

sec 21^{\circ} - sin 69^{\circ}

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