Solving a Homogeneous PDE with u_{xxx} - 3u_{xxy} + 4u_{yyy} = e^{x+2y}

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In summary, the correct approach to solving the given homogeneous equation is to factor out the common term $(dx - dy)$ and correctly simplify the expression to $(dx + dy)(dx - (2 + \sqrt{5})dy)(dx - (2 - \sqrt{5})dy)u = 0$.
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Homework Statement



[tex] u_{xxx} - 3u_{xxy} + 4u_{yyy} = e^{x+2y} [/tex]

The Attempt at a Solution



Ok so I tried doing the following to solve the homogeneous equation
\begin{align*}

u_{xxx} + u_{xxy} - 4u_{xxy} + 4_{yyy} = 0 \\

[d^{2}x(dx -dy) - 4dy(dx + dy)(dx -dy)]u = 0 \\

[(dx - dy)[d^{2}x - 4dxdy - 4d^{2}y]u = 0 \\

(dx -dy)(dx - (2+ /sqrt{5})dy)(dx - (2 - /sqrt{5})dy)u = 0

\end{align*}

Is this correct? Seems way too hard... I'm sure I did something incorrectly.
Thanks!
 
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  • #2


Your approach to solving the homogeneous equation is correct. However, the final expression in your attempt at a solution is not entirely accurate. The correct expression should be:

$(dx + dy)(dx - (2 + \sqrt{5})dy)(dx - (2 - \sqrt{5})dy)u = 0$

You have correctly factored out the common term $(dx - dy)$, but the second term should be $+dy$ instead of $-dy$. This is because when you expand the expression, you should have $dx^2 - (2 + \sqrt{5})dxdy + (2 + \sqrt{5})dxdy - (2 + \sqrt{5})(2 - \sqrt{5})dy^2$, which simplifies to $dx^2 - (2 + \sqrt{5})(2 - \sqrt{5})dy^2$. This is the same as $dx^2 - 4dy^2$, which is what you had in your expression.

Remember to always double check your algebra when solving equations. It's easy to make small mistakes, but they can greatly affect the final result. Keep up the good work!
 

1. What is a PDE?

A PDE, or partial differential equation, is an equation that involves partial derivatives of a multivariable function. It is commonly used in mathematics and physics to describe relationships between multiple variables.

2. How do you factor a PDE?

To factor a PDE, you can use techniques such as separation of variables, substitution, or integration by parts. The specific method used will depend on the form of the PDE and the given initial or boundary conditions.

3. Can all PDEs be factored?

No, not all PDEs can be factored. Some PDEs may have unique solutions that cannot be factored, while others may require advanced techniques such as numerical methods or series solutions.

4. What are some real-world applications of PDEs?

PDEs have a wide range of applications in various scientific fields, such as heat transfer, fluid dynamics, quantum mechanics, and population dynamics. They are also used in engineering for modeling and solving problems in areas like structural analysis and control systems.

5. Are there any software or tools available to help factor PDEs?

Yes, there are several software and tools available to help factor PDEs, such as MATLAB, Mathematica, and Maple. These programs have built-in functions and algorithms specifically designed for solving and manipulating PDEs.

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