# Homework Help: Linear Equations DE

1. Jan 27, 2007

### snowJT

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

I'm told that this is a linear equation

$$y' = \frac{4ln|x| - 2x^2y}{x^3}$$

2. The attempt at a solution

$$x^3 = (4ln|x|-2x^2y)\frac{dx}{dy}$$

giving:
$$P = 4ln|x|-2x^2$$
$$Q = x^3$$
$$\int Pdx = 4x-\frac{2x^3}{3}$$
$$e^\int^p^d^x = e^4^x^-^\frac^{2x^3}^{3}$$
$$ye^\int^p^d^x = \int e^4^x^-^\frac^{2x^3}^{3} x^3$$

I know this is too complicated and ugly.. It has to be wrong so far...

Last edited: Jan 27, 2007
2. Jan 27, 2007

### Dick

That is not a linear equation. Linear equations have the property that if y_1 and y_2 are solutions, then so is A*y_1+B*y_2. Your solution seems to consist of just throwing symbols around at random. The form of the equation doesn't make me think there is any way to get a closed form solution. If you really need to solve it you may have to try numerical techniques.

3. Jan 27, 2007

### snowJT

my textbook describes it as "first order linear DE"

I'm suppose to solve the equation by putting it into the form $$\frac{dy}{dx} + Py = Q$$

then integrate

4. Jan 27, 2007

### Dick

Ok. But then your assignments of P and Q don't look at all right. Concentrate of the form in 1) and try again.

5. Jan 27, 2007

### Dick

I see. It's an 'integrating factor' trick. Forgot.

6. Jan 27, 2007

### snowJT

is that what it's called? because I can not find anything on the internet that helps me with I look for liner equation DE

7. Jan 27, 2007

### Dick

Just go by your book. But you've identified P and Q wrong.

8. Jan 27, 2007

### HallsofIvy

That is, in fact, a linear equation for y as a function of x. Rewriting to
$$\frac{dx}{dy}$$ however gives you a non-linear equation for x as a function of y. There is no reason to do that.

The equation
$$\frac{dy}{dx}= \frac{4ln|x|-2x^2y}{x^3}$$
can be rewritten as
$$\frac{dy}{dx}+ \frac{2}{x}y= \frac{4 ln|x|}{x^3}$$

Now you are looking for an "integrating factor" $\mu(x)$ such that
$$\mu(x)\frac{dy}{dx}+ \mu(x)\frac{2}{x}y= \frac{d(\mu(x)y}{dx}$$
doing the derivative on the right, this reduces to
$$\mu(x)\frac{2}{x}y= \frac{d\mu(x)}{dx}y$$

In other words we must have
[tex]\frac{d\mu(x)}{dx}= \frac{2}{x}[/itex]
That's an easy separable equation.

Solve it to find $\mu(x)$ and the rest is easy.

9. Jan 27, 2007

### Dick

Mmm. I was waiting for him to identify the P and Q. Sorry about my confusing gaffe in saying it was not linear.

10. Jan 28, 2007

S'awright!