# Solve this partial diff. equation using substitution

## 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?

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fresh_42
Mentor
What is ##x'_x##?

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 :/

pasmith
Homework Helper
Is the PDE $$y\frac{\partial z}{\partial x} - x \frac{\partial z}{\partial y} = xyz$$

Your "relevant equations" is missing the chain rule: $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x}$$

• etotheipi
haruspex
Homework Helper
Gold Member
@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.

Last edited:
epenguin
Homework Helper
Gold Member
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?

haruspex