How to Solve a Partial Differential Equation with Variable Coefficients

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Homework Help Overview

The discussion revolves around solving a partial differential equation with variable coefficients, specifically the equation \(\frac{\partial{\psi}}{\partial{x}} = k(x+y)\). The original poster expresses difficulty in recalling methods for this type of equation and seeks guidance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration of the equation with respect to \(x\), treating \(y\) as a constant. There are questions about the correctness of terms in the solution and the impact of constants during integration.

Discussion Status

Some participants have offered insights into the integration process and the nature of the solution. There is acknowledgment of potential errors in calculations, and the original poster is encouraged to revisit their approach after identifying a mistake.

Contextual Notes

The original poster mentions feeling rusty with partial differential equations and expresses uncertainty about their previous notes on similar equations. There is a reference to a specific tutorial link, though its relevance is not discussed in detail.

Beer-monster
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Hi does anyone know how to solve this partial differential equation. My brain appears to be burping (and strangely my past notes don't seem to have any similar equation in):confused:
[tex]\frac{\partial{\psi}}{\partial{x}} = k(x+y)[/tex]
Anyone know any good tutorials or webpages for these sorts of equation? I'm a bit rusty with them
 
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[tex]\frac{\partial{\psi}}{\partial{x}} = k(x+y)[/tex]

Since x and y are independent, all you can do is integrate, with respect to x, treating y as a constant:
[tex]\psi(x,y)= \frac{k}{2}x^2+ yx+ f(y)[/tex]
Since the partial derivative wrt x is taken treating y as a constant, f(y) could be any function of y alone- its derivative will be 0.
 
I tried that but didn't get the right answer, I'll give it another shot.

Although isn't the second term kyx as it too is multiplied by k? Or am I being incredibly dense, it does happen a lot
 
Turn out the reason I was going wrong wasn't my method, I lost a minus in the calculation (slippery little blighters):rolleyes:

Thanks for the help anyway
 

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