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Math Help - solving by substituting a complex function

  1. #1
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    solving by substituting a complex function

    i need to solve
    x^2y''-xy'+y=4x^2

    by saying that x=e^t
    and Y(t)=y(x(t))

    then i need to find from them expresions for \fracxdy/dx and  x^2d^2y/dx^2 which are expresions of dY/dt similar type

    i tried like this:
    x=e^t
    \frac{dx}{dt}=e^t
    Y(t)=y(x(t))
    \frac{dY(t)}{dt}=\frac{dy(x)}{dx}\frac{dx(t)}{dt}
    \frac{dY(t)}{dt}=\frac{dy(x)}{dx}x
    \frac{dY^2(t)}{dt}=\frac{d^2y(x)}{dx^2}x^2+\frac{d  y(x)}{dx}

    y'x=\frac{dY(t)}{dt}
    y''x^2=\frac{dY^2(t)}{dt}-y'

    so
    \frac{dY^2(t)}{dt}-y'-\frac{dY(t)}{dt}+y=4x^2

    what now
    ?
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  2. #2
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    I'm not sure you're doing the second derivative properly. I agree with the first derivative, but you have to use the chain rule on the second derivative as well. See here.
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  3. #3
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    i replaces e^t with x
    so when i do the chain rules in the second derivative dx(t)/dt=e^t=x
    that how i got x^2
    i cant see the mistake
    in general i dont see how this method works
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  4. #4
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    You've correctly got

    \dot{Y}=\dfrac{dy}{dx}\,\dfrac{dx}{dt}=e^{t}y', and hence y'=e^{-t}\dot{Y}.

    Next, you compute

    \ddot{Y}=\dfrac{d}{dt}\left(e^{t}\,\dfrac{dy}{dx}\  right)<br />
=e^{t}\dfrac{dy}{dx}+e^{t}\dfrac{d}{dt}\,\dfrac{dy  }{dx}<br />
=\dot{Y}+e^{t}\left(\dfrac{d}{dx}\dfrac{dy}{dx}\ri  ght)\dfrac{dx}{dt}<br />
=\dot{Y}+e^{2t}y''.

    Thus,

    \ddot{Y}-\dot{Y}=e^{2t}y'', and therefore

    y''=e^{-2t}(\ddot{Y}-\dot{Y}).

    Substituting into your DE yields

    e^{2t}(e^{-2t}(\ddot{Y}-\dot{Y}))-e^{t}(e^{-t}\dot{Y})+Y=4e^{2t}.

    When you simplify, lots of nice things happen. Does this make sense?
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  5. #5
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    my prof said that x=e^t
    and when i see dx/dt i replace it with x
    because dx/dt=e^t=x

    so there isnt supposed to be e^t in your solution
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  6. #6
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    Well, continue simplifying, and substitute back in so you just have x's. It's six of one, a half-dozen of the other.

    Incidentally, this method of solution is inferior to the usual Cauchy-Euler method, because with this solution, you are artificially constraining x to be positive, whereas that is not the case with the usual Cauchy-Euler method.
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  7. #7
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    (\ddot{Y}-x^2\dot{Y}))-(\dot{Y})+xY=4x^2
    \ddot{Y}+(-x^2-1)\dot{Y}+xY=4x^2

    ok i did the substitution
    how it became easier
    ?
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  8. #8
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    That is incorrect. How did you get that from my post # 4?
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  9. #9
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    <br />
 e^{2t}(e^{-2t}(\ddot{Y}-\dot{Y}))-e^{t}(e^{-t}\dot{Y})+Y=4e^{2t} <br />
e^t=x <br />

    <br />
e^2t=x^2<br /> <br />
    4 members canceled each other
    (\ddot{Y}-x^2\dot{Y}))-(\dot{Y})+xY=4x^2
    \ddot{Y}+(-x^2-1)\dot{Y}+xY=4x^2
    i cant see my mistake

    and its still a second order equation with x's as coefficient
    i cant see how it got easier
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  10. #10
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    You made two mistakes. The first one was in not noticing that the e^{-2t} multiplies both terms in the expression \ddot{Y}-\dot{Y}. The second mistake is the mysterious x that multiplies the Y as your last term on the LHS. That seems to come out of thin air. The e^{t} multiplying the e^{-t}\dot{Y} term does NOT multiply the Y term.

    Perhaps both of these mistakes are a result of not noticing the parentheses carefully enough? You absolutely have to pay attention to details like parentheses if you're going to avoid mistakes.
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  11. #11
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    ok here i go carefull on every step
    first i open the parentheses and put x^2 instead of e^{2t}
    \ddot{Y}-2\dot{Y}+Y=4x^2

    but Y is a function of t
    not x
    how to go further
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  12. #12
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    Correct. Now, here's where I would differ with your professor, perhaps. You should have the exponentials on the RHS, not x^2. Then you've got a second-order linear non-homogeneous ode with constant coefficients. Turn the crank.
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  13. #13
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    ok but how this method differs the y=x^a substitution method
    ?

    my prof said that the exponent method is better because it solves equations that
    x^a cant

    in wikipedea they dont say anything about the difference between them
    Cauchy eiler link

    on what type i need to use what method
    ?
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  14. #14
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    I'm not overly familiar with the differences between the two methods, other than the observation I made before. If you have a bona-fide Cauchy-Euler equation, then the x^a method should solve it. The exponential substitution might work for other types of DE's, but as I said, you're restricting yourself to positive x's. That might not matter so much if your x is really a t, and you're only interested in positive time.
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  15. #15
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    cant understand the transition from
    e^{t}\dfrac{d}{dt}\,\dfrac{dy}{dx}
    to
    e^{t}\left(\dfrac{d}{dx}\dfrac{dy}{dx}\right)\dfra  c{dx}{dt}<br />
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