Solving linear differential equation

[ArcanaNoir]

Homework Statement

I'm trying to make sense of the steps for solving a linear differential equation and I'm stuck at the part where
\mu (x) \frac{\mathrm{d}y}{\mathrm{d}x} + \mu (x) P(x)y=\frac{\mathrm{d}}{\mathrm{d}x} [\mu (x)y]

I guess I'm not sure what \frac{\mathrm{d}}{\mathrm{d}x} even means. Doesn't it mean the derivative with respect to x?

Here's an example, perhaps that will help to clarify the trouble.
x^{-2} \frac{\mathrm{d}y}{\mathrm{d}x} - 2x^{-3}y is supposedly equivalent to
\frac{\mathrm{d}}{\mathrm{d}x} (x^{-2}y)

Now how does that equate?

Homework Equations

The Attempt at a Solution

The derivative of (x^{-2}y) is - 2x^{-3}y.
So... why doesn't \frac{\mathrm{d}}{\mathrm{d}x} (x^{-2}y) = - 2x^{-3}y ?

 

[rock.freak667]

d/dx here is the total differential.

So for d/dx(x^-2y), x and y are both variables. So you'd apply the product law here i.e. u = x^-2 and v=y.

 

[I like Serena]

Happy new year, ArcanaNoir!

In your equation you have ${dy \over dx}$.
This confirms that y is a function of x. that is, y=y(x).

When you determine the derivative $\frac{\mathrm{d}}{\mathrm{d}x} (x^{-2}y)$, you need to apply the product rule for derivatives.

 

[ArcanaNoir]

What is the difference between d/dx and dy/dx?

 

[I like Serena]

d/dx is the derivative with respect to x of the function that comes after.

dy/dx is the derivative of y=y(x) with respect to x.

 

[ArcanaNoir]

How come the derivative of the d/dx expression pops out with a dy/dx?

 

[I like Serena]

Suppose you had to differentiate $x^{-2}g(x)$, what would the derivative be?

 

[ArcanaNoir]

-2x^(-3)g(x)+x^(-2)g'(x)

Ah, so since y is a function of x, aha. Thank you a lot for helping me dissect this.