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Paritial derivative.

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Verify if ##t=\lambda x## then ##x^2\frac{\partial^2 y}{\partial x^2} = t^2\frac{\partial^2 y}{\partial t^2}##

    3. The attempt at a solution

    [tex]t=\lambda x\;\Rightarrow\; \frac{\partial t}{\partial x}=\lambda[/tex]
    [tex]\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t}\frac{\partial t}{\partial x}= \lambda\;\frac{\partial y}{\partial t}[/tex]
    [tex]\frac{\partial^2 y}{\partial x^2}=\lambda \frac{\partial^2 y}{\partial t^2} \frac{\partial t}{\partial x}=\lambda^2\frac{\partial^2 y}{\partial t^2}[/tex]

    [tex]x^2\frac{\partial^2 y}{\partial x^2}=\frac{t^2}{\lambda^2}\lambda^2\frac{\partial^2 y}{\partial t^2}\;\Rightarrow\;x^2\frac{\partial^2 y}{\partial x^2} = t^2\frac{\partial^2 y}{\partial t^2}[/tex]

    Am I correct?

  2. jcsd
  3. Oct 1, 2013 #2


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    Yes, with just a minor quibble:
    "t" is only dependent on "x", and not in addition dependent on other variables.
    Thus, you should use dt/dx, rather than the symbol for the partial derivative here.
  4. Oct 1, 2013 #3


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    Your differential equation has dependent variable "y" so it makes no sense to ask if "[itex]t= \lambda x[/itex]" satisfies the equation.

    I suspect you are asked to show that [itex]f(t-\lambda x)[/itex] satisfies the equation for f any twice differentiable function of a single variable.
  5. Oct 1, 2013 #4


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    I don't agree HallsofIvy. Here, it is about how a simple linear scaling will affect the shape of a particular derivative from the "unscaled" x as independent variable, to its shape in its "scaled" variable "t". That is, we are dealing with the correct IDENTITY, not an equation!! It is not about an "equation" as such, but meant as an intermediate step in transforming an equation from the "x"-formulation to its logically equivalent "t"-formulation.

    TECHNICALLY, we are looking at the relation between the shapes of the derivatives of functions y and Y, respectively, fulfilling the relation

    making an abuse of notation with replacing "Y" with "y" everywhere.
    Last edited: Oct 1, 2013
  6. Oct 1, 2013 #5
    Thanks everyone. I meant exactly what I asked, it's only about the steps to perform the differential equation.
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