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Math Help - PDE - Yet another

  1. #1
    Member Altair's Avatar
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    PDE - Yet another

    This is my third question regarding PDE solvable as ODE and I am still confused because the book I am following (Advanced Engineering Mathematics by Erwin Kreyszig) has only two very elementary examples where as I find the exercise problems to be very difficult, especially in their approach.

    Currently I am having problem in Q5 of Problem Set 12.1.

    u_y + u = e^{xy}

    How do I approach this one. I write u_y in differential form but then I do not know how to handle the e^{xy} part. Can someone please explain the solutiion to this one.

    And please can I be provided with some links to online examples for PDEs solvable as ODEs ?
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    Quote Originally Posted by Altair View Post
    This is my third question regarding PDE solvable as ODE and I am still confused because the book I am following (Advanced Engineering Mathematics by Erwin Kreyszig) has only two very elementary examples where as I find the exercise problems to be very difficult, especially in their approach.

    Currently I am having problem in Q5 of Problem Set 12.1.

    u_y + u = e^{xy}

    How do I approach this one. I write u_y in differential form but then I do not know how to handle the e^{xy} part. Can someone please explain the solutiion to this one.

    And please can I be provided with some links to online examples for PDEs solvable as ODEs ?
    Not sure on links to PDEs that can be solved as ODEs but I'll comment on the one you have here. You'll notice that your PDE has no x derivative so x is treated as a constant in your PDE. This problem is now like the following ODE

     <br />
u' + u = e^{cy}<br />

    It's linear and can be solved using an integrating factor. Remember, put back the x( c \to x) and with the constant of integration, it's a fucntion of integration.
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    A perhaps daft question to tag on the end here, but i thought the integrating factor method would only work for "proper" derivatives, not partial derivatives? I don't know where ive got this idea from, perhaps someone could clarify?
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    Quote Originally Posted by OnkelTom View Post
    A perhaps daft question to tag on the end here, but i thought the integrating factor method would only work for "proper" derivatives, not partial derivatives? I don't know where ive got this idea from, perhaps someone could clarify?
    The key thing here is the absence of the x derivative. So the integrating factor for

    u_y + u is \mu = e^{\int 1dy} = e^y

    and sure enough

     <br />
e^y\left(u_y+u\right) = \left(e^y u\right)_y<br />
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    MHF Contributor chisigma's Avatar
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    Since the partial derivative respect to x is absent , we can consider the equation written as...

    u^{'} + u = e^{x\cdot y} (1)

    ... where the independent variable is y and x is a constant. The (1) is a standard linear DE of first order and its solution is...

    u(y) = c\cdot e^{-y} + \varphi(y) (2)

    ... where \varphi(*) is a particular solution of (1) that is found solving an indefinite integral...

    \varphi(y)= e^{-y}\cdot \int e^{(1+x)x\cdot y}\cdot dy = \frac{e^{x\cdot y}}{1+x} (3)

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    Since the partial derivative respect to x is absent , we can consider the equation written as...

    u^{'} + u = e^{x\cdot y} (1)

    ... where the independent variable is y and x is a constant. The (1) is a standard linear DE of first order and its solution is...

    u(y) = c\cdot e^{-y} + \varphi(y) (2)

    ... where \varphi(*) is a particular solution of (1) that is found solving an indefinite integral...

    \varphi(y)= e^{-y}\cdot \int e^{(1+x)x\cdot y}\cdot dy = \frac{e^{x\cdot y}}{1+x} (3)

    Kind regards

    \chi \sigma
    I think you mean \varphi(x,y)= \frac{e^{x\cdot y}}{1+x}. What you have doesn't make sense. Further the general solution of the PDE is

    u = \frac{e^{x\cdot y}}{1+x} + f(x) e^{-y} where f(x) is arbitrary (not a constant as you have it).
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    MHF Contributor chisigma's Avatar
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    Excuse me sir!...

    ... explain me please what is the difference in the two cases...

    a)  u = \frac {e^{x\cdot y}}{1+x} + c \cdot e^{-y} where x and c are two constants independent from y

    b) u = \frac {e^{x\cdot y}}{1+x} + f(x) \cdot e^{-y} where x is an arbitrary constant independent from y and f(x) is an arbitrary function of the constant x

    ... a little mysterious question ...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    Excuse me sir!...

    ... explain me please what is the difference in the two cases...

    a)  u = \frac {e^{x\cdot y}}{1+x} + c \cdot e^{-y} where x and c are two constants independent from y

    b) u = \frac {e^{x\cdot y}}{1+x} + f(x) \cdot e^{-y} where x is an arbitrary constant independent from y and f(x) is an arbitrary function of the constant x

    ... a little mysterious question ...

    Kind regards

    \chi \sigma
    Sure. u = \frac {e^{x\cdot y}}{1+x} + f(x) \cdot e^{-y} is much more general than u = \frac {e^{x\cdot y}}{1+x} + c \cdot e^{-y} and contains it as a special case (i.e. where f(x) = c)

    Side note, x is not a constant but a variable as y is.
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  9. #9
    MHF Contributor chisigma's Avatar
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    ... I understand ...

    Kind regards

    \chi \sigma
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    Member Altair's Avatar
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    Quote Originally Posted by danny arrigo View Post
    The key thing here is the absence of the x derivative. So the integrating factor for

    u_y + u is \mu = e^{\int 1dy} = e^y

    and sure enough

     <br />
e^y\left(u_y+u\right) = \left(e^y u\right)_y<br />
    Can you please explain this method...As to how do I identify that what is going to be the I.F and what is the geenral form of this method and how do I know when to apply this ?
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  11. #11
    Member Altair's Avatar
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    Quote Originally Posted by chisigma View Post
    Since the partial derivative respect to x is absent , we can consider the equation written as...

    u^{'} + u = e^{x\cdot y} (1)

    ... where the independent variable is y and x is a constant. The (1) is a standard linear DE of first order and its solution is...

    u(y) = c\cdot e^{-y} + \varphi(y) (2)

    ... where \varphi(*) is a particular solution of (1) that is found solving an indefinite integral...

    \varphi(y)= e^{-y}\cdot \int e^{(1+x)x\cdot y}\cdot dy = \frac{e^{x\cdot y}}{1+x} (3)

    Kind regards

    \chi \sigma
    Same question goes for you chisigma
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  12. #12
    Member Altair's Avatar
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    The kind I have solved with I.F are given in the following tutorial under point 3.

    \frac{dy}{dx}+P(x)y=Q(x)<br />

    Where as in this question (which I posted) there isn't an Q(x) only on the right side of equality sign. Its e^{xy}. So even if x is constant its still a Q(y) and not Q(x)

    http://www.mathhelpforum.com/math-he...-tutorial.html
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    Quote Originally Posted by Altair View Post
    The kind I have solved with I.F are given in the following tutorial under point 3.

    \frac{dy}{dx}+P(x)y=Q(x)

    Where as in this question (which I posted) there isn't an Q(x) only on the right side of equality sign. Its e^{xy}. So even if x is constant its still a Q(y) and not Q(x)

    http://www.mathhelpforum.com/math-he...-tutorial.html
    Your PDE (an ODE now that we're setting x constant)

    \frac{du}{dy} + u = e^{cy}\;\;(1)

    Comparing with your form

     <br />
\frac{dy}{dx} + P(x)y = Q(x)\;\;(2)<br />

    all we need to do is identify the variables going from (2) to (1) (y is u and x is y).
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