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gerryliyana

  • 3 years ago

Differential Equations again!

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  1. gerryliyana
    • 3 years ago
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    SOLVE \[\frac{ dy }{ dx } + y = xy^{3}\]

  2. UnkleRhaukus
    • 3 years ago
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    you have a Bernoulli equation \[y'+p(x)y=q(x)y^n\], subsitute \(z=y^{1-n}\)

  3. gerryliyana
    • 3 years ago
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    Ok..., thank u @UnkleRhaukus So.., \[\frac{ dy }{ dx } + y = xy^{3}\] \(y^{-3} \frac{ dy }{ dx } +y^{-2} = x\) ........(1) \(z = y^{1-n} \) where n = 3, then \(z = y^{-2}\) \[\frac{ dz }{dx } = \frac{ dz }{ dy }\frac{ dy }{ dx } = \frac{ d }{ dy }(y^{-2}) \frac{ dy }{ dx } = -2 y^{-3} \frac{ dy }{ dx }\] \[\frac{ dy }{ dx } = \frac{ 1 }{ -2y } y^{3} \frac{ dz }{ dx }\] substitute dy/dx into the (1) equations, becoms \[y^{-3} (\frac{ 1 }{ -2 }) y^{3} \frac{ dz }{ dx } + y^{-2} = x\] \(\frac{ dz }{ dx } -2y^{-2} = -2x\) where \(z = y^{-2}\) \(\frac{ dz }{ dx } -2z = -2x\) ....(2) i got p(x) = -2 and g(x) = -2x Then for integration factor: \[\mu(x) = \exp(\int\limits_{0}^{x} -2 dx) = \exp(-2x) = e^{-2x}\] multiply all (2) terms by \( \mu (x)\) , becomes \[e^{-2x} \frac{ dz }{ dx } -2e^{-2x}z = -2e^{-2x} x\] \[d(e^{-2x} z) = -2 e^{-2x} x\] \[e^{-2x} z = -2 \int\limits_{0}^{x} e^{-2x} x dx\] that is right??

  4. gerryliyana
    • 3 years ago
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    \[e^{-2x} z = \frac{ 1 }{ 2 }e^{-2x} (2x+1) + C\] where z = y^{-2} \[e^{-2x} y^{-2} = \frac{ 1 }{ 2 } e^{-2x}(2x+1) + C\]

  5. UnkleRhaukus
    • 3 years ago
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    thats right

  6. UnkleRhaukus
    • 3 years ago
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    now just rearrange and solve for y(x)

  7. gerryliyana
    • 3 years ago
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    wow..., very nice.., ok., \[e^{-2x}y^{-2} = \frac{ 1 }{ 2 } e^{-2x} (2x+1) + C\] \[e^{-2x}y^{-2} - \frac{ 1 }{ 2 } e^{-2x}(2x+1) = 0\] \[y^{-2} - \frac{ 1 }{ 2 } (2x+1) = 0\] \[y^{2} = \frac{ 2 }{ (2x+1) } + C\] \[y = \sqrt{\frac{ 2 }{ (2x+1)}+ C}\] am i right????

  8. UnkleRhaukus
    • 3 years ago
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    almost , the term withe the constant is not quite right though , something strange happend on your second line just now ^

  9. gerryliyana
    • 3 years ago
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    I'm a little bit confused about C

  10. UnkleRhaukus
    • 3 years ago
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    \[e^{-2x}y^{-2} - \frac{ 1 }{ 2 } e^{-2x}(2x+1) - C=0\] \[y^{-2} - \frac{ 1 }{ 2 } (2x+1) - Ce^{2x}=0\]

  11. UnkleRhaukus
    • 3 years ago
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    \[y^{-2}= \frac{ 1 }{ 2 }( (2x+1) + 2Ce^{2x})\] \[y^{2}= \frac{ 2 }{ (2x+1) + 2Ce^{2x}}\] \[y^{2}(x)= \frac{ 2 }{ (2x+1) + ke^{2x}}\] \[y(x)= \pm\sqrt{\frac{ 2 }{ (2x+1) + ke^{2x}}}\]

  12. UnkleRhaukus
    • 3 years ago
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    maybe just leave the solution as \[y^{2}(x)\]

  13. UnkleRhaukus
    • 3 years ago
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    \[\qquad2C=k\]

  14. gerryliyana
    • 3 years ago
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    ok..., thank u so much :)

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