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Particular Solutions of Differential Equations

  • Thread starter paul2211
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  • #1
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Homework Statement



[itex]\frac{d^{2}y}{dx^{2}}[/itex] = [itex]y\frac{dy}{dx}[/itex]

Homework Equations



Let [itex] v = \frac{dy}{dx}[/itex] and [itex]v\frac{dv}{dy} = \frac{d^{2}y}{dx^{2}}[/itex]


The Attempt at a Solution



The question can be rewritten as:

[itex]v\frac{dv}{dy} = yv[/itex]
[itex]\frac{dv}{dy} = y[/itex]. (v =/=0 )

This is very easy to solve since it's basically a normal integral. I get v and substitute in [itex]\frac{dy}{dx}[/itex] to get an implicit expression for y:

[itex]C+\frac{x}{2}= D Tan^{-1}(Dy)[/itex]

However, the problem is when I divided v on both sides, and I noted that v can't be 0 because division by 0 is not allowed.

Thus, v = 0 is a particular solution to the DE, so y equals a constant is not a solution to this DE?

I really hope someone can give me a better understanding of particular solutions, and what I should do with them in a problem such as this.

Thank you very much.
 

Answers and Replies

  • #2
HallsofIvy
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Thus, v = 0 is a particular solution to the DE, so y equals a constant is not a solution to this DE?
No, that says just the opposite! v= 0 is a solution to the equation after you had "reduced" it so y equal to a constant is a solution to the DE.

If y is a constant, both first and second derivatives are 0 so your equation just becomes 0= y(0).
 

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