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Math Help - General Solution for the characteristics of 2 quasi linear PDE's

  1. #1
    Senior Member bugatti79's Avatar
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    General Solution for the characteristics of 2 quasi linear PDE's

    Hi all,

    I am stuck on how the general solutions where obtained for the following quasi linear PDE's. I understand this to be quasi linear because the highest derivative of u is linear even though e^u is nonlinear.
    Anyway,
    1) yu_x-xu_y=e^u and 2) xu_x+yu_y=sec(u)

    I calculate the characteristics of 1) and 2) as
    1) \frac{dx}{dt}=y, \frac{dy}{dt}=-x, \frac{du}{dt}=e^u
    hence \frac{dy}{dx}=\frac{-x}{y} however I dont know how this arrived at x=acos(t) + bsin(t), y=bcos(t)-asin(t), -e^{-u}=t+c
    How are there 2 constants a and b when it is only a first order ODE?

    2) \frac{dx}{dt}=x, \frac{dy}{dt}=y, \frac{du}{dt}=sec(u)
    hence \frac{dy}{dx}=\frac{y}{x} which arrives at
    x=Ae^t, y=Be^t, sin(u)=t+c

    I dont know how x,y and t+c are determined?
    Any tips will be appreciated.
    Thanks
    bugatti79
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  2. #2
    MHF Contributor
    Jester's Avatar
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    On the first since

    \dfrac{dx}{dt} = y,\;\;\; \dfrac{dy}{dt} = -x then \dfrac{d^2 x}{dt^2} = \dfrac{dy}{dt} = -x so

    \dfrac{d^2 x}{dt^2} + x = 0 which has the solution

    x = a \cos t + b \sin t and

    y = \frac{dx}{dt} = - a \sin t + b \cos t.

    On the second, integrate

    \dfrac{dx}{dt} = x,\;\;\; \dfrac{dy}{dt} = y,\;\;\;\dfrac{du}{dt} = \sec u separately
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    On the first since

    \dfrac{dx}{dt} = y,\;\;\; \dfrac{dy}{dt} = -x then \dfrac{d^2 x}{dt^2} = \dfrac{dy}{dt} = -x so

    \dfrac{d^2 x}{dt^2} + x = 0 which has the solution

    x = a \cos t + b \sin t and

    y = \frac{dx}{dt} = - a \sin t + b \cos t.

    On the second, integrate

    \dfrac{dx}{dt} = x,\;\;\; \dfrac{dy}{dt} = y,\;\;\;\dfrac{du}{dt} = \sec u separately
    It was right in front of my eyes!!

    so \dfrac{d^2 x}{dt^2} + x = 0 is a standard 2nd order ODE


    \int e^{-u}du=\intdt yields -e^{-u} = t +c

    \frac{dx}{dt}=x yields ln x = t+c
    x=e^{t+c}=e^{t} e^{c}=Ae^{t} likewise for dy/dt = y


    Thanks Danny
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