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Astrophysics

  • one year ago

Reduction of order (To find second solution)

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  1. Astrophysics
    • one year ago
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    I have to use the method of reduction order to find a second solution to this equation \[t^2y''+2ty'-2y=0, ~~~t>0;y_1(t)=t\] I checked this solution it checks out but the one I got...\[y(t)=v(t)y_1(t)=v(t)t\] \[y'(t)=v'(t)t+v(t)\] and \[y''(t)=v''(t)t+v'(t)\]

  2. Astrophysics
    • one year ago
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    I set the equation up as \[y'' + \frac{ 2 }{ t }y'-\frac{ 2 }{ t^2 }y=0\] and then I plugged in above \[v''(t)t+v'(t)+\frac{ 2 }{ t } \left[ v'(t)t+v(t) \right]-\frac{ 2 }{ t^2 }v(t)t=0 \implies v''(t)t+2v'(t)=0\] then I made the substitutions v'(t)=u(t), and v''(t)=u'(t) making my equation \[u'(t)t+3u(t)=0 \implies u't+3u=0\]\[u=t^{-3}+C_1~~~~C_1=e^C\] now subbing back v'(t)=u(t) \[v'=t^{-3}+C_1 \implies v(t) = -\frac{ t^{-2} }{ 2 }+C_2\]

  3. Astrophysics
    • one year ago
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    But the back of the book says the second solution is \[y_2 = t^{-2}\]

  4. anonymous
    • one year ago
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    I actually figured out a bit of a simpler way to solve for a secondary solution for second order differential equation with nonconstant coefficients. See here: http://openstudy.com/study#/updates/561b2901e4b07ab19da73807 The professor showed using \[y_2 = y_1 \int\limits \frac{ e^{- \int\limits p(t) dt} }{ y_1^2 }dt\]

  5. Astrophysics
    • one year ago
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    I would do that, but we haven't learnt it haha

  6. anonymous
    • one year ago
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    Does your professor want you to strictly use the method they taught? If not, this might save time. =/ I haven't learned how to solve with nonconstant coefficients yet, so I can't exactly help. But from the other thread I learned how to solve using that method.

  7. Astrophysics
    • one year ago
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    Yeah, we have to use this haha

  8. anonymous
    • one year ago
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    Ah, darn. Sorry :(

  9. Astrophysics
    • one year ago
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    It's cool! Thanks though!

  10. Astrophysics
    • one year ago
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    \[y_1(t) = t~~~y_1'(t)=1~~~y_1''(t)=0\] uhh

  11. dan815
    • one year ago
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    cauchy form

  12. dan815
    • one year ago
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    try t^n solution

  13. Astrophysics
    • one year ago
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    No man I can't use that

  14. dan815
    • one year ago
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    o ok

  15. ShadowLegendX
    • one year ago
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    i lag

  16. freckles
    • one year ago
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    \[t^2 y''+2t y'-2y=0 \\ t>0 \\ y_1(t)=t \\ \text{ assume } y_2(t)=v \cdot y_1=tv \\ y_2'=tv'+v \\ y_2''=v'+tv''+v'=tv''+2v' \\ \text{ plugging in } \\ t^2(t v''+2 v')+2t(t v'+v)-2(tv)=0 \\ t^3 v''+2t^2v'+2t^2v'+2tv-2tv=0 \\ t^3 v''+4t^2 v'=0 \\ t>0 \\ \text{ so we have } \\ v''+4v'=0 \]

  17. dan815
    • one year ago
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    how come this always happens

  18. freckles
    • one year ago
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    you have 2 instead of 4 there

  19. dan815
    • one year ago
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    well now thats just a first order ode with a sub

  20. freckles
    • one year ago
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    \[\text{ oops type-0 } \\ t v''+4v'=0\]

  21. Astrophysics
    • one year ago
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    Oh you know what I realized my second derivative is wrong as well wow

  22. Astrophysics
    • one year ago
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    Ooh ok I see freckles, thanks a million!

  23. dan815
    • one year ago
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    like variation of parameter thing if we are plugging in a real solution the lowest order v should always drop out?

  24. freckles
    • one year ago
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    \[u=v' \\ t u'+4u=0 \\ t \frac{du}{dt}=-4 u \\ \frac{du}{u} =\frac{-4}{ t} dt \\ \ln|u|=-4 \ln|t|+k \\ \ln|u|=-4 \ln|t|+\ln|C| \\ \ln|u|=\ln|t^{-4} C| \\ u=t^{-4} C \\ v'=t^{-4} C\]

  25. Astrophysics
    • one year ago
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    Yes, that should work :), thanks @freckles

  26. freckles
    • one year ago
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    np lol for some reason when I did it on paper I made the same mistake you did but I was like whatever made I made a mistake so I just started over typing here

  27. Astrophysics
    • one year ago
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    Haha, yeah I made it twice actually :x eek, I usually find the mistake when I write it on here as well

  28. freckles
    • one year ago
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    actually no I just didn't know how to add 2 and 2 for some reason

  29. Astrophysics
    • one year ago
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    xD

  30. freckles
    • one year ago
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    I did the differentiating right but for some reason when I got to 2+2 I put 3 down for some reason

  31. Astrophysics
    • one year ago
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    Haha silly arithmetic, thanks again!

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