Solving Linear Equation: y' = 4ln|x|-2x^2y/x^3

[snowJT]

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...

 

[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.

 

[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

 

[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.

 

[Dick]

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

 

[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

 

[Dick]

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

 

[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
\frac{d\mu(x)}{dx}= \frac{2}{x}[/itex]<br /> That&#039;s an easy separable equation.<br /> <br /> Solve it to find \mu(x) and the rest is easy.

 

[Dick]

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

 

[HallsofIvy]

S'awright!