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ironictoaster

  • 2 years ago

ODE problem confused!

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  1. ironictoaster
    • 2 years ago
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  2. ironictoaster
    • 2 years ago
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    Confused on b(ii)

  3. ironictoaster
    • 2 years ago
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    I'm not sure what to do.

  4. ironictoaster
    • 2 years ago
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    I have to find u somehow I reckon, not 100% though.

  5. Spacelimbus
    • 2 years ago
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    Hmm I believe you just have to show that the identity is valid, because the roots are identical.

  6. colorful
    • 2 years ago
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    pretty similar to the part before it

  7. Spacelimbus
    • 2 years ago
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    But I will try first.

  8. Spacelimbus
    • 2 years ago
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    the other solution will have an additional x infront of it I am not mistaken.

  9. colorful
    • 2 years ago
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    juts plug it in and see what happens

  10. experimentX
    • 2 years ago
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    http://en.wikipedia.org/wiki/Reduction_of_order

  11. ironictoaster
    • 2 years ago
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    If it's double roots then yes there's extra x in the solution.

  12. ironictoaster
    • 2 years ago
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    I'm still not sure what should I do first.

  13. ironictoaster
    • 2 years ago
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    I don't know what u is.

  14. experimentX
    • 2 years ago
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    assume that \( x(t) = u(t) x_1(t) \) is another solution ... find the value of u(t) ... so that you have complete solution.

  15. ironictoaster
    • 2 years ago
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    Still lost...

  16. ironictoaster
    • 2 years ago
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    @experimentX I genuinely don't know where to start.

  17. experimentX
    • 2 years ago
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    i think there is an example in the wikipedia ... in the link i posted above.

  18. Spacelimbus
    • 2 years ago
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    If I understand this problem then they just want you to check what happens if you substitute back their provided result. I believe their are trying to introduce you to the method of Reduction of Order You will get a result in the form of \[ \Large u(x)=d_1x+d_2 \] where \(d_1, d_2\) are constant. The second solution is of the form \[ \Large y_2(x)=u(x)e^{\frac{x}{2}}\] So you can use superposition to get the general solution.

  19. experimentX
    • 2 years ago
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    |dw:1344360084786:dw|

  20. Spacelimbus
    • 2 years ago
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    Note that \[ \Large 4r^2-4r+1=0 \] Has a discriminant of zero.

  21. Spacelimbus
    • 2 years ago
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    Pardon me if I was interrupting something in here, OpenStudy lags horribly for me today so I hit the post button before it crashes me again (-:

  22. ironictoaster
    • 2 years ago
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    and how does that prove it equals 0?

  23. ironictoaster
    • 2 years ago
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    Yeah it pretty bad this week

  24. ironictoaster
    • 2 years ago
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    What I understand so far, we need to sub ux into equation 4

  25. Spacelimbus
    • 2 years ago
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    \[4\left( u''e^{\frac{t}{2}} + \frac{1}{2}e^{\frac{t}{2}}u' + \frac{1}{4}e^{\frac{t}{2}}u + \frac{1}{2}u'e^{\frac{t}{2}}\right) -4 \left(u' e^{\frac{t}{2}}+ \frac{1}{2}e^{\frac{t}{2}}u \right)+ ue^{\frac{t}{2}}=0\]

  26. Spacelimbus
    • 2 years ago
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    Divide by \( \large e^{\frac{t}{2}} \) and then see what happens when you bring it all together.

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