# Homework Help: Solving linear differential equation

1. Jan 15, 2012

### ArcanaNoir

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

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?

2. Relevant equations

3. The attempt at a solution

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

2. Jan 15, 2012

### 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.

3. Jan 15, 2012

### 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.

4. Jan 15, 2012

### ArcanaNoir

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

5. Jan 15, 2012

### 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.

6. Jan 15, 2012

### ArcanaNoir

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

7. Jan 15, 2012

### I like Serena

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

8. Jan 15, 2012

### 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.