(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I'm working through the classic treatise on ODEs by Ince. I know that this is a somewhat dated text, but (imho) there are some real "gems" on the subject of ODEs here, well worthy of careful study.

I'm looking at Problem #7 at the end of Chapter 1. In it, we are given a function [tex]z(x,y)[/tex]:

[tex]

z(x,y) = 3xy - y^2 + (y^2-2x)^{3/2}

[/tex]

and we are asked to show two things. First, that:

[tex]

\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x}

[/tex]

and second that:

[tex]

\frac{\partial^2 z}{\partial x^2} \cdot \frac{\partial^2 z}{\partial y^2} = \left( \frac{\partial^2z}{\partial x \partial y} \right)^2

[/tex]

The first one is a no-brainer. It's the second one that I'm having trouble with..

2. Relevant equations

All are given above.

3. The attempt at a solution

Differentiating the above expression, we have:

[tex]

z_x = 3y + \left( \frac{3}{2} \right)\left(y^2-2x \right)^{1/2}(-2)

[/tex]

[tex]

z_x = 3y - 3\left(y^2- 2x\right)^{1/2}

[/tex]

and differentiating again:

[tex]z_{xy} = 3 - 3\left(\frac{1}{2}\right) \left(y^2-2x\right) ^ {-1/2}(2y)[/tex]

[tex]z_{xy} = 3 - \frac{3y}{\sqrt{y^2-2x}}[/tex]

Now differentiating with respect to y first, we have:

[tex]z_y = 3x - 2y + \left(\frac{3}{2}\right) \left(y^2-2x\right)^{1/2}(2y)[/tex]

[tex]z_y = 3x - 2y + 3y\sqrt{y^2-2x}[/tex]

and differentiating again:

[tex]z_{yx} = 3 +3y\left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(-2)[/tex]

[tex]z_{yx} = 3 - \frac{3y}{\sqrt{y^2-2x}}[/tex]

So yes, we have [tex]z_{yx} = z_{xy}[/tex] as expected.

Now, moving on to the second part of this problem, we square this expression:

[tex]\left(\frac{\partial^2z}{\partial x \partial y}\right)^2 = 9 + \frac{9y^2}{y^2-2x} - \frac{18y}{\sqrt{y^2-2x}}[/tex]

[tex]\left(\frac{\partial^2z}{\partial x \partial y}\right)^2 = \frac{18(y^2-x)}{y^2-2x} - \frac{18y}{\sqrt{y^2-2x}}[/tex]

Now, going back to the expression for [tex]z_x[/tex]:

[tex]z_x = 3y - 3\sqrt{y^2-2x}[/tex]

and differentiating again with respect to x, we have:

[tex]z_{xx} = -3\left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(-2)[/tex]

[tex]z_{xx} = \frac{3}{\sqrt{y^2-2x}}[/tex]

And going back to the expression for [tex]z_y[/tex]:

[tex]z_y = 3x - 2y + 3y\sqrt{y^2-2x}[/tex]

and differentiating with respect to y:

[tex]z_{yy} = -2 + 3\left[\sqrt{y^2-2x} + y \left(\frac{1}{2}\right)\left(y^2-2x\right)^{-1/2}(2y)\right][/tex]

[tex]z_{yy} = - 2 + 3 \left[\sqrt{y^2-2x} + \frac{y^2}{\sqrt{y^2-2x}}\right][/tex]

[tex]z_{yy} = - 2 + 3 \left[\frac{2(y^2-x)}{\sqrt{y^2-2x}}\right] [/tex]

[tex]z_{yy} = - 2 + \frac{6(y^2-x)}{\sqrt{y^2-2x}}[/tex]

Multiplying these together, we get:

[tex]z_{xx} \cdot z_{yy} = \frac{18(y^2-x)}{y^2-2x}- \frac{6}{\sqrt{y^2-2x}}[/tex]

Comparing this to what we derived above for [tex]\left(z_{xy}\right)^2[/tex], we see it's close, but not exact. The first term is correct, but the second term is off by a factor of 3y.

What am I doing wrong?

Or is there a typo in Ince?

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# Homework Help: Ince's ODEs

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