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!