# Linear Equations DE

1. Homework Statement

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...

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Dick
Homework Helper
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.

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

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

Dick
Homework Helper
I see. It's an 'integrating factor' trick. Forgot.

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

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

HallsofIvy
Homework Helper
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.

Dick