The solution of the differential equation x 2 -y x 2 dy dx - y 2 -x y 2 -0 is

The correct option is A log( x y ) = 1 x + 1 y +C Given differential equation is ( x ^ 2 y x ^ 2 ) dy dx + y ^ 2 +x y ^ 2 =0 ⇒ 1 y ...

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Question

The solution of the differential equation $(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$ is

log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C

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log (\frac{y}{x}) = \frac{1}{x} + \frac{1}{y} + C

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log (x y) = \frac{1}{x} + \frac{1}{y} + C

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log (x y) + \frac{1}{x} + \frac{1}{y} = C

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Similar questions

(1 + x) y d x + (1 + y) x d y = 0

log x y + (x - y) = c .

y = e^{x} + 1

y^{''} - y = 0

y = x^{2} + 2 x + C

y^{'} - 2 x - 2 = 0

y = cos x + C

y^{'} + sin x = 0

y = \sqrt{1 + x^{2}}

y^{'} = \frac{x y}{1 + x^{2}}

y = A x

x y = y (x \neq 0)

y = x sin x

x y = y + x \sqrt{x^{2} y^{2}} (x \neq 0 a n d x > y o r x < y)

x y = log y + C

y^{'} = \frac{y^{2}}{1 - x y} (x y \neq 1)

y - cos y = x

(y sin y + cos y + x) y^{'} = y

x + y = {tan}^{- 1} y

y^{2} y^{'} + y^{2} + 1 = 0

y = \sqrt{a^{2} - x^{2}} x ϵ (- a, a)

x + y \frac{d y}{d x} = 0 (y \neq 0)

\frac{d y}{y} + \frac{d x}{x} = 0

The general solution of the differential equation $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$ is

x, y, z

x + log (x + \sqrt{x^{2} + 1}) = y

y + log (y + \sqrt{y^{2} + 1}) = z

z + log (z + \sqrt{z^{2} + 1}) = x

{(1 - x)}^{2} + {(1 - y)}^{2} + {(1 - z)}^{2}

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