Help rearranging a linear first order differential equation

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SUMMARY

The discussion focuses on rearranging the linear first-order differential equation dy/dx = 3x^2 - 2x + 2 + (8/x * y) into standard form. The correct rearrangement involves isolating the y terms, resulting in the equation dy/dx - (8/x)y = 3x^2 - 2x + 2. This form allows for the application of an integrating factor, μ(x) = exp(∫P(x)dx), to solve the equation. Participants confirmed the clarity of the rearrangement process and expressed confidence in completing the solution.

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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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