Help rearranging a linear first order differential equation

[Kris1]

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 :(

 

[Kris1]

Re: Help rearranging a linear first order differential

The equation of interest is the first of the two you have listed. Thankyou :)

 

[Chris L T521]

Re: Help rearranging a linear first order differential

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!

 

[Kris1]

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 :)

 

[blue_raver22]

Sure, let's break down the steps to rearrange this linear first order differential equation:

1. First, let's group all the terms with y on one side of the equation and all the terms with x on the other side. This will give us:

dy/dx - (8/x) * y = 3x^2 - 2x + 2

2. Next, we need to factor out the coefficient of y, which in this case is -8/x. This will give us:

dy/dx - (8/x) * y = (3x^2 - 2x + 2)

3. Now, we need to divide both sides of the equation by the coefficient of y, -8/x. This will give us:

(dy/dx - (8/x) * y) / (-8/x) = (3x^2 - 2x + 2) / (-8/x)

4. Simplifying the left side of the equation, we get:

dy/dx - y = (3x^2 - 2x + 2) / (-8/x)

5. Finally, we can rearrange the terms to get the equation in standard form, which is:

dy/dx + y = (2x^2 - 2x + 2) / (8/x)

And there you have it! The linear first order differential equation has been rearranged to standard form. I hope this helps clear up any confusion. Remember, when solving differential equations, it's important to follow the rules of algebra and manipulate the equation to get it in a form that is easier to work with.