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Linear Equations DE

  • Thread starter snowJT
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1. Homework Statement

I'm told that this is a linear equation

[tex]y' = \frac{4ln|x| - 2x^2y}{x^3}[/tex]

2. The attempt at a solution

[tex]x^3 = (4ln|x|-2x^2y)\frac{dx}{dy}[/tex]

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

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

Answers and Replies

Dick
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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.
 
118
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my textbook describes it as "first order linear DE"

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

then integrate
 
Dick
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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
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I see. It's an 'integrating factor' trick. Forgot.
 
118
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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
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Just go by your book. But you've identified P and Q wrong.
 
HallsofIvy
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That is, in fact, a linear equation for y as a function of x. Rewriting to
[tex]\frac{dx}{dy}[/tex] however gives you a non-linear equation for x as a function of y. There is no reason to do that.

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

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

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 [itex]\mu(x)[/itex] and the rest is easy.
 
Dick
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Mmm. I was waiting for him to identify the P and Q. Sorry about my confusing gaffe in saying it was not linear.
 
HallsofIvy
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S'awright!
 

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