# Thread: Having problems with a linear 1st order ODE

1. ## Having problems with a linear 1st order ODE

I've got a linear 1st order ODE, which I'm trying to solve via integrating factors:

$\displaystyle \frac{dy}{dx} = 3 - 6x + y - 2xy$

$\displaystyle y = -3 + ce^{x^2 - x}$

So, from here, this is what I start out with:

$\displaystyle \frac{dy}{dx} + y(-1 + 2x) = 3 - 6x$

$\displaystyle P(t) = 2x -1$

$\displaystyle U = e^{\int 2x - 1 dx} = e^{x^2 - x}$

$\displaystyle \int \frac{d}{dx}[ye^{x^2 - x}] = \int 3e^{x^2 - x} - 6xe^{x^2 - x} dx$

But working from here would seem to require integrating $\displaystyle e^{x^2 - x}$, which I didn't think was trivially doable.

Where am I going wrong?

2. ## Re: Having problems with a linear 1st order ODE

Originally Posted by Lancet
I've got a linear 1st order ODE, which I'm trying to solve via integrating factors:

$\displaystyle \frac{dy}{dx} = 3 - 6x + y - 2xy$

$\displaystyle y = -3 + ce^{x^2 - x}$

So, from here, this is what I start out with:

$\displaystyle \frac{dy}{dx} + y(-1 + 2x) = 3 - 6x$

$\displaystyle P(t) = 2x -1$

$\displaystyle U = e^{\int 2x - 1 dx} = e^{x^2 - x}$

$\displaystyle \int \frac{d}{dx}[ye^{x^2 - x}] = \int 3e^{x^2 - x} - 6xe^{x^2 - x} dx$

But working from here would seem to require integrating $\displaystyle e^{x^2 - x}$, which I didn't think was trivially doable.

Where am I going wrong?
Put x=x'+2, dx=dx' in the ODE.

Edit: will not work!

3. ## Re: Having problems with a linear 1st order ODE

Originally Posted by Also sprach Zarathustra
Put x=x'+2, dx=dx' in the ODE.
I don't understand. I also don't get why this can be done...

4. ## Re: Having problems with a linear 1st order ODE

Originally Posted by Lancet
I've got a linear 1st order ODE, which I'm trying to solve via integrating factors:

$\displaystyle \frac{dy}{dx} = 3 - 6x + y - 2xy$

$\displaystyle y = -3 + ce^{x^2 - x}$

So, from here, this is what I start out with:

$\displaystyle \frac{dy}{dx} + y(-1 + 2x) = 3 - 6x$

$\displaystyle P(t) = 2x -1$

$\displaystyle U = e^{\int 2x - 1 dx} = e^{x^2 - x}$

$\displaystyle \int \frac{d}{dx}[ye^{x^2 - x}] = \int 3e^{x^2 - x} - 6xe^{x^2 - x} dx$

But working from here would seem to require integrating $\displaystyle e^{x^2 - x}$, which I didn't think was trivially doable.

Where am I going wrong?
If you split up your integral you will have problems. Consider your integral (I have factored out a 3)

$\displaystyle \int 3(1-2x)e^{x^2-x}dx$

and try the substitution $\displaystyle u = x^2-x$.

BTW, your initial ODE is separable

$\displaystyle \dfrac{dy}{dx} = 3 - 6x + y - 2xy = (y+3)(1-2x)$

5. ## Re: Having problems with a linear 1st order ODE

Originally Posted by Danny
If you split up your integral you will have problems. Consider your integral (I have factored out a 3)

$\displaystyle \int 3(1-2x)e^{x^2-x}dx$

and try the substitution $\displaystyle u = x^2-x$.

BTW, your initial ODE is separable

$\displaystyle \dfrac{dy}{dx} = 3 - 6x + y - 2xy = (y+3)(1-2x)$
How I didn't see that?!

6. ## Re: Having problems with a linear 1st order ODE

Originally Posted by Danny

If you split up your integral you will have problems. Consider your integral (I have factored out a 3)

Wow, that's a nasty integration example, simple though it is. Thank you for showing that to me!

Originally Posted by Danny

BTW, your initial ODE is separable