Solve the following differential equations:
$x d x + y d y = x d y - y d x$

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

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $ln (x^{2} + y^{2}) = 2 {tan}^{- 1} (\frac{y}{x}) + c$
$x d x + y d y = x d y - y d x \Rightarrow y \frac{d y}{d x} + x = x \frac{d y}{d x} - y$
Substituting $y = x v \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
$x + x (x \frac{d v}{d x} + v) v = x (x \frac{d v}{d x} + v) - x v \Rightarrow x + x (x \frac{d v}{d x} + v) v = x^{2} \frac{d v}{d x} \Rightarrow \frac{d v}{d x} = \frac{- v^{2} - 1}{x (v - 1)} \Rightarrow \frac{\frac{d v}{d x} (v - 1)}{- v^{2} - 1} = \frac{1}{x}$
Integrating both sides w.r.t $x$ , we get
$\int \frac{\frac{d v}{d x} (v - 1)}{- v^{2} - 1} d x = \int \frac{1}{x} d x \Rightarrow {tan}^{- 1} v - \frac{1}{2} log (v^{2} + 1) = log x + c {tan}^{- 1} \frac{y}{x} - \frac{1}{2} log ({(\frac{y}{x})}^{2} + 1) = log x + c$

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Similar questions

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}

If log (x^{2} + y^{2}) = 2 {tan}^{- 1} (\frac{y}{x}),

\frac{d y}{d x} = \frac{x + y}{x - y} .

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

\frac{d y}{d x} =

y^{2} d x + (x^{2} - x y + y^{2}) d y = 0

\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})

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