1. ## PDE Help

Ok, my question deals with the following PDE:

Dxx(u) + 3*Dyy(u) - 2*Dx(u) + 24*Dy(u) +5*u = 0

and there is a substitution I am instructed to use:

u = v*e^(a*x+By), and y' = d*y. Where a, B, and d are constants

I am trying to reduce the PDE down to the form Dmm(v) + Dnn(v) + c*v = 0

I don't understand what to do with the d I. have done the substitution, and and off by a factor of 3 on the Dnn(v), so I have determined that with 2 partials that d must be 1/sqrt(3), however I don't understand how it works in the problem. I know it fixes my factor of 3 problem, but I don't understand how it works.

The form at the moment is of:

Dmm(v)*e^(ax+By) +3*Dnn(v)*e^ax+By) + (a-1)*Dm(v)*e^(ax+By) + (6B+24)*Dn(v)*e^(ax+By) +(a^2 + B^2 + 24B - 2a +5)*v*e^(ax+By)

So I have determined also that a = 1 and B = -4.... But how does d come in in order to get rid of the factor of 3?

BTW sorry about not using easier notation, wasn't familiar with how it works on the forum.

2. Consider the equation of the ellipse

$x^2 + \dfrac{y^2}{3} = 1$

If we introduce new variables

$x^{*} = x,\;\;\; y^{*} = \dfrac{y}{\sqrt{3}}\;\;\;(1)$

the ellipse becomes the circle

$x^{*}^2 + y^{*}^2 = 1.$

The change of variables (1) stretches the ellipse to a circle (or vice-verse). This change of variable acts the same way transforming

$u_{xx} + 3 u_{yy} + lots = 0$

to

$u_{x^{*}x^{*}} + u_{y^{*}y^{*}}} + lots = 0$

3. Awesome, things make sense now. Thanks again!