Help rearranging a linear first order differential equation

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Discussion Overview

The discussion revolves around rearranging a linear first-order differential equation of the form dy/dx = 3x^2 - 2x + 2 + (8/x * y). Participants seek clarification on how to manipulate the equation into standard form for solving.

Discussion Character

  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant expresses confusion about rearranging the differential equation into standard form.
  • Another participant provides a step-by-step approach to isolate terms involving y, suggesting to subtract (8/x)y from both sides to achieve the form dy/dx - (8/x)y = 3x^2 - 2x + 2.
  • The same participant notes that the resulting equation fits the linear form dy/dx + P(x)y = Q(x) and mentions the need to compute the integrating factor, leaving that part for the original poster to complete.
  • A later reply indicates that the original poster has gained clarity on the rearrangement process and feels more confident in proceeding with the solution.

Areas of Agreement / Disagreement

Participants generally agree on the method to rearrange the equation, but the discussion does not resolve the complete solution process, as one participant leaves out the calculation of the integrating factor.

Contextual Notes

The discussion does not address specific assumptions regarding the function forms or the domain of x, nor does it resolve the steps for calculating the integrating factor.

Kris1
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Hi I am trying to solve dy/dx = 3x^2-2x+2+(8/x *y)

Can anyone show me how to rearrange to standard form as I am mightly confused :(
 
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Re: Help rearranging a linear first order differential

The equation of interest is the first of the two you have listed. Thankyou :)
 
Re: Help rearranging a linear first order differential

Kris said:
Hi I am trying to solve dy/dx = 3x^2-2x+2+(8/x *y)

Can anyone show me how to rearrange to standard form as I am mightly confused :(

The first thing you need to do is get every term involving y on one side of the equation. So subtracting $\dfrac{8}{x}y$ from both sides gives you
\[\frac{dy}{dx}-\frac{8}{x}y=3x^2-2x+2.\]
We now note that the equation is now in the form of a linear equation $\dfrac{dy}{dx}+P(x)y=Q(x)$. To proceed from here, you need to compute the integrating factor
\[\mu(x)=\exp\left(\int P(x)\,dx\right)=\ldots\quad(\text{I leave this part to you})\]
where $\exp(x)=e^x$. Then if you multiply both sides of the linear ODE by $\mu(x)$, you get
\[\frac{d}{dx}[\mu(x) y]=\mu(x)(3x^2-2x+2)\implies y=\frac{1}{\mu(x)}\int \mu(x)(3x^2-2x+2)\,dx.\]

Can you fill in the work I left out? I hope this helps!
 
Re: Help rearranging a linear first order differential

Yes thanks I can fill the rest out I was just unsure as how to rearrange the equation because of all the terms multiplied by x but I see that it is quite easy now :)
 

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