Homework Help: Particular Solutions of Differential Equations

1. Nov 16, 2011

paul2211

1. The problem statement, all variables and given/known data

$\frac{d^{2}y}{dx^{2}}$ = $y\frac{dy}{dx}$

2. Relevant equations

Let $v = \frac{dy}{dx}$ and $v\frac{dv}{dy} = \frac{d^{2}y}{dx^{2}}$

3. The attempt at a solution

The question can be rewritten as:

$v\frac{dv}{dy} = yv$
$\frac{dv}{dy} = y$. (v =/=0 )

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

$C+\frac{x}{2}= D Tan^{-1}(Dy)$

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.

2. Nov 17, 2011

HallsofIvy

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).