Solve this partial diff. equation using substitution

  • Thread starter Addez123
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  • #1
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Homework Statement:

$$yx'_x - xz'_y = xyz$$
Solve by variable change:
$$u = x^2 + y^2$$
$$v = e ^{-x^2/2}$$

Relevant Equations:

None.
I completely forgot how to solve these so heres my attempt:
$$z = au + bv$$
$$z = a(x^2 + y^2) + be ^{-x^2/2}$$
$$z'_x = 2ax - bxe ^{-x^2/2}$$
$$z'_y = 2ay$$
Put that into the original equation and you get
$$y * (2ax - bxe ^{-x^2/2}) -x * (2ay) = $$
$$-ybe^{-x^2/2} = xyz$$
$$z = -be^{-x^2/2}/x$$

But the solution should be
$$e^{-x^2/2} * f(x^2 + y^2)$$

Did I do everything wrong or just missed something here?
 

Answers and Replies

  • #3
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What is ##x'_x##?
If I extract x from u = x^2 + y^2 I get:
$$x = sqrt(u - y^2)$$
and getting the derivative requires me to replace u, which leads to x = x so idk :/
 
  • #4
pasmith
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Is the PDE [tex]
y\frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = xyz[/tex]

Your "relevant equations" is missing the chain rule: [tex]
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x}[/tex]
 
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  • #5
haruspex
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@Addez123 , Why do you think that particular change of variables is helpful for this equation?
I had more success writing the equation as ##\frac{z_x}x-\frac{z_y}y=z##, then reasoning that I needed solutions like ##\frac{z_x}x=Az## etc.
 
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  • #6
epenguin
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Addez has not answered the question asked (#2) which, put another way is: it is not clear what the question is and its formula looks to be a mistranscription so can we be informed/confirmed what our problem is?
 
  • #7
haruspex
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Addez has not answered the question asked (#2) which, put another way is: it is not clear what the question is and its formula looks to be a mistranscription so can we be informed/confirmed what our problem is?
It is obvious enough that ##x_x## is a typo for ##z_x##, and that the primes in ##z'_x## etc. are redundant. In post #1, @Addez123 differentiated the expression substituted for z partially wrt x and wrote the result as ##z'_x=...##.
 

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