Respuesta:
[tex]z = \frac{y}{x} [/tex]
[tex](2 {x}^{3} + {y}^{3} )dx = 3x {y}^{2} dy[/tex]
[tex] \frac{2 {x}^{3} + {y}^{3} }{3x {y}^{2} } = \frac{dy}{dx} [/tex]
[tex]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} [/tex]
[tex]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[/tex]
[tex]x \frac{dz}{dx} = \frac{2}{ {3z}^{2} } - \frac{2z}{3} \rightarrow \: x \frac{dz}{dx} = \frac{2 - 2 {z}^{3} }{3 {z}^{2} } [/tex]
[tex] - \frac{1}{2} \int - 2 \frac{3 {z}^{2} }{2 - {2z}^{3} } dz = \int \frac{1}{x} dx[/tex]
Sea: u = 2 - 2z³ ; du = -6z²dz
[tex] - \frac{1}{2} \int \frac{1}{x} du = ln(x) + C \rightarrow \: - \frac{1}{2} ln(u) = ln(x) + C[/tex]
[tex] - \frac{1}{2} ln(2 - 2 {z}^{3} ) = ln(x) + C [/tex]
[tex] ln(2 - {2z}^{3} ) ^{ - \frac{1}{2} } = ln(x) + C[/tex]
[tex]{e}^{ ln(2 - {2z}^{3} ) ^{ - \frac{1}{2} } } = {e}^{ ln(x) + C} [/tex]
[tex](2 - {2z}^{3} ) ^{ - \frac{1}{2} } = {e}^{ ln(x) } . {e}^{C} = xD[/tex]
[tex](2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2}[/tex]
[tex](2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2} \rightarrow \: \frac{1}{2 - {2z}^{3} } =E {x}^{2} [/tex]
[tex]1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3})}{x} [/tex]
[tex]1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3}) }{x} [/tex]
[tex] \frac{1x}{2E} = x^{3} - {y}^{3} [/tex]
[tex]F = {x}^{3} - {y}^{3} [/tex]
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Verified answer
Respuesta:
[tex]z = \frac{y}{x} [/tex]
[tex](2 {x}^{3} + {y}^{3} )dx = 3x {y}^{2} dy[/tex]
[tex] \frac{2 {x}^{3} + {y}^{3} }{3x {y}^{2} } = \frac{dy}{dx} [/tex]
[tex]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} [/tex]
[tex]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[/tex]
[tex]x \frac{dz}{dx} = \frac{2}{ {3z}^{2} } - \frac{2z}{3} \rightarrow \: x \frac{dz}{dx} = \frac{2 - 2 {z}^{3} }{3 {z}^{2} } [/tex]
[tex] - \frac{1}{2} \int - 2 \frac{3 {z}^{2} }{2 - {2z}^{3} } dz = \int \frac{1}{x} dx[/tex]
Sea: u = 2 - 2z³ ; du = -6z²dz
[tex] - \frac{1}{2} \int \frac{1}{x} du = ln(x) + C \rightarrow \: - \frac{1}{2} ln(u) = ln(x) + C[/tex]
[tex] - \frac{1}{2} ln(2 - 2 {z}^{3} ) = ln(x) + C [/tex]
[tex] ln(2 - {2z}^{3} ) ^{ - \frac{1}{2} } = ln(x) + C[/tex]
[tex]{e}^{ ln(2 - {2z}^{3} ) ^{ - \frac{1}{2} } } = {e}^{ ln(x) + C} [/tex]
[tex](2 - {2z}^{3} ) ^{ - \frac{1}{2} } = {e}^{ ln(x) } . {e}^{C} = xD[/tex]
[tex](2 - {2z}^{3} ) ^{ - \frac{1}{2} } = {e}^{ ln(x) } . {e}^{C} = xD[/tex]
[tex](2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2}[/tex]
[tex](2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2} \rightarrow \: \frac{1}{2 - {2z}^{3} } =E {x}^{2} [/tex]
[tex](2 - {2z}^{3} ) ^{ - 1} = {x}^{2} {D}^{2} \rightarrow \: \frac{1}{2 - {2z}^{3} } =E {x}^{2} [/tex]
[tex]1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3})}{x} [/tex]
[tex]1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3}) }{x} [/tex]
[tex]1 = E {x}^{2} (2 - \frac{2 {y}^{3} }{ {x}^{3} }) \rightarrow \: 1 = \frac{2E( {x}^{3} - {y}^{3}) }{x} [/tex]
[tex] \frac{1x}{2E} = x^{3} - {y}^{3} [/tex]
[tex]F = {x}^{3} - {y}^{3} [/tex]