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ksaimouli

  • one year ago

find particular solution y=f(x

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  1. goformit100
    • one year ago
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    its easy

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

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

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

  5. Spacelimbus
    • one year ago
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    Integrate both sides.

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

  7. ksaimouli
    • one year ago
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    @Spacelimbus

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

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

  10. ksaimouli
    • one year ago
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    no calc

  11. ksaimouli
    • one year ago
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    calculator

  12. Spacelimbus
    • one year ago
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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.

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

  14. Spacelimbus
    • one year ago
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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.

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

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

  17. ksaimouli
    • one year ago
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    @Spacelimbus

  18. ksaimouli
    • one year ago
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    i alway get these

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

  20. ksaimouli
    • one year ago
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    okay

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

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