- #1

McCoy13

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- 0

## Homework Statement

Given the differential equation

[tex]u_{xx}+3u_{yy}-2u_{x}+24u_{y}+5u=0[/tex]

use the substitution of dependent variable

[tex]u=ve^{ \alpha x + \beta y}[/tex]

and a scaling change of variables

[tex]y'= \gamma y[/tex]

to reduce the differential equation to

[tex]v_{xx}+v_{yy}+cv=0[/tex]

## Homework Equations

I have no idea

## The Attempt at a Solution

I tried a direct substitution of both variables:

[tex]u_{x}=\alpha ve^{\alpha x+\beta y}[/tex]

[tex]u_{xx}=\alpha^{2}ve^{\alpha x+\beta y}[/tex]

[tex]u_{y}=\beta ve^{\alpha x+\beta y}[/tex]

[tex]u_{yy}=\beta^{2} ve^{\alpha x+\beta y}[/tex]

Plugging in this gives

[tex]\alpha^{2} ve^{\alpha x+\beta y}+3\beta^{2}ve^{\alpha x+\beta y}-2\alpha ve^{\alpha x+\beta y}+24\beta ve^{\alpha x+\beta y}+5ve^{\alpha x+\beta y}[/tex]

You can obviously factor out [itex]ve^{\alpha x+\beta y}[/itex], but that doesn't really do much for you. I also tried doing this with the y' substitution. It also occurred to me that since v is probably supposed to be understood as v(x,y), I tried this set of substitutions:

[tex]u_{x}=\alpha ve^{\alpha x+\beta y}+v_{x}e^{\alpha x+\beta y}[/tex]

[tex]u_{xx}=\alpha^{2} ve^{\alpha x+\beta y}+\alpha v_{x}e^{\alpha x+\beta y}+v_{xx}e^{\alpha x+\beta y}[/tex]

[tex]u_{x}=\beta ve^{\alpha x+\beta y}+v_{y}e^{\alpha x+\beta y}[/tex]

[tex]u_{xx}=\beta^{2} ve^{\alpha x+\beta y}+\beta v_{y}e^{\alpha x+\beta y}+v_{yy}e^{\alpha x+\beta y}[/tex]

None of these attempts gave me any insight into the problem.

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