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Thread: 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) $\displaystyle yu_x-xu_y=e^u$ and 2) $\displaystyle xu_x+yu_y=sec(u)$

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

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

    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
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    On the first since

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

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

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

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

    On the second, integrate

    $\displaystyle \dfrac{dx}{dt} = x,\;\;\; \dfrac{dy}{dt} = y,\;\;\;\dfrac{du}{dt} = \sec u$ 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

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

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

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

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

    On the second, integrate

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

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


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

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


    Thanks Danny
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