# Homework Help: First order linear diff eq

1. May 15, 2012

### robertjford80

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

3. The attempt at a solution

I don't understand in step one why the three in the numerator disappears. I also don't understand why dy/dx becomes d/dx. the book just says left side is d/dx(v*y), a lot of help that is. how do you go to two fractions with x^3 and x^4 to one fraction with x^3 in the numerator.

In step 3 they integrate both sides, but I don't see any integration of the left side. I have a feeling that they are moving the d/dx to the right side, if so, what happens to the d?

2. May 15, 2012

### sharks

It's a first-order linear ODE, and usually, the method of solving this is by multiplying throughout by an integrating factor.

3. May 15, 2012

### robertjford80

I know that but i still don't understand how they get from steps 1 to 2 to 3

4. May 15, 2012

### sharks

I confirm. It's exactly the same method, except that some of the intermediary steps have been omitted from your example.

You should re-arrange the linear ODE into its standard form. Then, find the integrating factor.

5. May 15, 2012

### robertjford80

well i don't know what those intermediary steps are. if i did i wouldn't have posted the question.

6. May 15, 2012

### sharks

The solution will reveal itself if you work it out and you will see the corresponding lines from your example.

To standardize your linear ODE, multiply by $x^3$:
$$\frac{dy}{dx}+\left(\frac{-3}{x}\right)y=x$$ Can you continue from here?

7. May 15, 2012

### robertjford80

I don't see how that step is legal. number two, I don't see how that gets me from step 1 to step 2.

8. May 15, 2012

### sharks

I thought you knew how to solve a first-order linear ODE.

Continuing from my post #6. From the standardized linear ODE form:
$$P(x)=\frac{-3}{x} \\Q(x)=x$$
The integrating factor:
$$\mu (x)=e^{\int P(x).dx}=\frac{1}{x^3}$$
$$\frac{d(\mu y)}{dx}=\mu Q$$
$$\frac{d(\frac{1}{x^3} y)}{dx}= \frac{1}{x^2}\; ......\;Step\;2$$
Steps 3 and 4 are obvious.

9. May 15, 2012

### Steely Dan

There's nothing omitted from the first step. It's simply a fact that if you take the derivative of the left hand side of the second equation, you get the left hand side of the first equation. Take the derivative (using the product rule) to confirm this.

10. May 15, 2012

### robertjford80

I'm having trouble taking the derivative of

$$\frac{1}{x^3} - \frac{3}{x^4}y$$

I get

$$\frac{dy}{dx}\big(\frac{x-3}{x^4}\big)y$$

If I take the derivative of that I'm not going to get the next step in the book, so I must be doing something wrong.

Last edited: May 15, 2012
11. May 15, 2012

### HallsofIvy

Well, that's your problem- you do NOT take the derivative of that. You want to integrate it.

I get

$$\frac{dy}{dx}\big(\frac{x-3}{x^3}\big)y$$

If I take the derivative of that I'm not going to get the next step in the book, so I must be doing something wrong.[/QUOTE]
The derivative of $y/x^3$, using the quotient rule, is
$$\frac{y'x^3- y(3x^2)}{(x^3)^2}= \frac{y'}{x^3}- \frac{3y}{x^4}[/quit] which is exactly what you want. 12. May 15, 2012 ### robertjford80 i realize this is basic algebra but I don't see to get [tex]\frac{y}{x^3}$$

If you let y = 2 and x = 2 you get

$$\frac{1}{x^3} - \frac{3y}{x^4}$$

$$\frac{2}{16} - \frac{3*2}{16} = \frac{-2}{8}$$

whereas with

$$\frac{y}{x^3}$$

if you let y = 2 and x = 2 you get

$$\frac{2}{8}$$

There's something about the negative sign that I'm missing.