Question

$(2 {x}^{3} + {y}^{3} )dx - 3x {y}^{2} dy = 0$
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Answer 1

Haha c-mamo con su nombre si usuario ‘Diosteamo7’ xdd

Answer 2

Respuesta:

$z = \frac{y}{x}$

$(2 {x}^{3} + {y}^{3} )dx = 3x {y}^{2} dy$

$\frac{2 {x}^{3} + {y}^{3} }{3x {y}^{2} } = \frac{dy}{dx}$

$y' = x \frac{dz}{dx} + z = \frac{ {x}^{3} }{3x {y}^{2} } + \frac{ {y}^{3} }{3x {y}^{2} } = \frac{ {2x}^{2} }{3 {y}^{2} } + \frac{y}{3x}$

$x \frac{dz}{dx} + z = \frac{2}{ {3z}^{2} } + \frac{z}{3} \rightarrow \: x \frac{dz}{dx} = \frac{2}{ {3z}^{2} } + \frac{z}{3} - z$

$x \frac{dz}{dx} = \frac{2}{ {3z}^{2} } - \frac{2z}{3} \rightarrow \: x \frac{dz}{dx} = \frac{2 - 2 {z}^{3} }{3 {z}^{2} }$

$- \frac{1}{2} \int - 2 \frac{3 {z}^{2} }{2 - {2z}^{3} } dz = \int \frac{1}{x} dx$

Sea: u = 2 - 2z³ ; du = -6z²dz

$- \frac{1}{2} \int \frac{1}{x} du = ln(x) + C \rightarrow \: - \frac{1}{2} ln(u) = ln(x) + C$

$- \frac{1}{2} ln(2 - 2 {z}^{3} ) = ln(x) + C$

$ln(2 - {2z}^{3} ) ^{ - \frac{1}{2} } = ln(x) + C$

${e}^{ ln(2 - {2z}^{3} ) ^{ - \frac{1}{2} } } = {e}^{ ln(x) + C}$

$(2 - {2z}^{3} ) ^{ - \frac{1}{2} } = {e}^{ ln(x) } . {e}^{C} = xD$

$(2 - {2z}^{3} ) ^{ - \frac{1}{2} } = {e}^{ ln(x) } . {e}^{C} = xD$

$(2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2}$

$(2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2} \rightarrow \: \frac{1}{2 - {2z}^{3} } =E {x}^{2}$

$(2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2} \rightarrow \: \frac{1}{2 - {2z}^{3} } =E {x}^{2}$

$1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3})}{x}$

$1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3}) }{x}$

$1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3}) }{x}$

$\frac{1x}{2E} = x^{3} - {y}^{3}$

$F = {x}^{3} - {y}^{3}$