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ksaimouli

find particular solution y=f(x

  • one year ago
  • one year ago

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  1. goformit100
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    its easy

    • one year ago
  2. ksaimouli
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    \[\frac{ dy }{ dx }=6x^2-x^2y\]

    • one year ago
  3. Spacelimbus
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    \[\Large \frac{dy}{dx}=(6-y)x^2 \] Or \[ \Large \frac{1}{6-y}dy=x^2dx \] For \(y(x) \neq 6\)

    • one year ago
  4. ksaimouli
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    \[\int\limits_{}^{}\frac{ 1 }{ 6-y }dy=x^2dx\]

    • one year ago
  5. Spacelimbus
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    Integrate both sides.

    • one year ago
  6. ksaimouli
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    \[-lny=2x^3+c\]

    • one year ago
  7. ksaimouli
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    @Spacelimbus

    • one year ago
  8. Spacelimbus
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    \[\Large - \ln |6-y(x)|=\frac{1}{3}x^3+C\prime  \]

    • one year ago
  9. ksaimouli
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    they have given f(-1)=2

    • one year ago
  10. ksaimouli
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    no calc

    • one year ago
  11. ksaimouli
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    calculator

    • one year ago
  12. Spacelimbus
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    You know you have to solve it for y(x) if you can, of course - you could apply the initial conditions just now and solve for the constant. But I strongly recommend you to get the equation in explicit form if possible.

    • one year ago
  13. ksaimouli
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    i got \[y(x)=ce ^{(x^3/x)}+6\]

    • one year ago
  14. Spacelimbus
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    Almost, try again. You should end up with: \[\Large y(x)=6-Ce^{-\frac{1}{3}x^3} \] As you can see, I said above that \(y(x) \neq 6\) which was important, otherwise I would have divided the differential equation through 0. This equation supports that statement, this equation only becomes 6 in the limit as x approaches infinity.

    • one year ago
  15. ksaimouli
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    |dw:1360441531205:dw|

    • one year ago
  16. ksaimouli
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    |dw:1360441583641:dw|

    • one year ago
  17. ksaimouli
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    @Spacelimbus

    • one year ago
  18. ksaimouli
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    i alway get these

    • one year ago
  19. Spacelimbus
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    Consider this equation: \[\Large e^{\ln2} = 2 \] but: \[\Large e^{-\ln2} = -2 ????? \] Rather view it like that: \[\Large e^{-\ln2}=\left(e^{\ln2}\right)^{-1}=\frac{1}{2} \]

    • one year ago
  20. ksaimouli
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    okay

    • one year ago
  21. Spacelimbus
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    and then you should get the same expression as I do.

    • one year ago
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