The solution of $x^{3} d y - y^{3} d x = 0$ is:

x^{4} - y^{4} = c

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x^{2} - y^{2} = c

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\frac{1}{x} - \frac{1}{y} = c

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

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Solution

The correct option is D $\frac{1}{x^{2}} - \frac{1}{y^{2}} = c$
$x^{3} d y - y^{3} d x = 0$

$\Rightarrow \int \frac{d y}{y^{3}} = \int \frac{d x}{x^{3}} + c$

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

$\Rightarrow \frac{1}{x^{2}} - \frac{1}{y^{2}} = c$

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Similar questions

\frac{2}{π} \oint_{c} (- y^{3} d x + x^{3} d y)

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

{\frac{1}{x} - \frac{y^{2}}{(x - y)^{2}}} d x + {\frac{x^{2}}{(x - y)^{2}} - \frac{1}{y}} d y = 0

c

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}

= [- 1, 1]

= [0, 1]

= [- 1, 0]

S_{1} = {(x, y) | x^{2} + y^{2} = 1, x ϵ A, y ϵ A}

S_{2} = {(x, y) | x^{2} + y^{2} = 1, x ϵ A, y ϵ B}

S_{3} = {(x, y) | x^{2} + y^{2} = 1, x ϵ A, y ϵ C}

S_{4} = {(x, y) | x^{2} + y^{2} = 1, x ϵ B, y ϵ C}

\frac{d y}{d x} + x (x + y) = x^{3} (x + y)^{3} - 1

^{'} C^{'}

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