differential equations

**topsquark** · July 11th 2007, 07:17 AM

Originally Posted by emily28

i dont that i was meant to look at particular solutions and expenituals, is there another way to do it using perhaps a different method, as i havent covered this in class and dont think the teacher would want me to approach it this way?

Probably Jhevon's integrating factor approach would be the best then.

-Dan

**emily28** · July 11th 2007, 08:16 AM

done a little on using the intergrating factor, but how would be the best way to get the formula into a form to use the intergrating formula?

**emily28** · July 11th 2007, 11:34 AM

does anyone have any suggestions/

**Krizalid** · July 11th 2007, 11:38 AM

Originally Posted by emily28

$\frac{du}{dx} + 4 u = -4x$

Find an integrating factor defined by

$\mu(x)=\exp\left(4\int~dx\right)$

**topsquark** · July 11th 2007, 12:30 PM

Originally Posted by emily28

does anyone have any suggestions/

See here.

-Dan

**Jhevon** · July 11th 2007, 07:00 PM

Originally Posted by emily28

does anyone have any suggestions/

I will say what Krizalid said, but in a more general way.

For a differential equation of the form: $y' + p(x)y = f(x)$

we define $\mu (x) = e^{\int p(x)~dx} = \mbox { exp} \left( \int p(x)~dx \right)$ as the integrating factor of the differential equation. We solve the differential equation by multiplying through by the integrating factor. so we get the new equation:

$\mu (x) y' + \mu (x) p(x) y = \mu (x) f(x)$

it will always be the case that the left hand side is the derivative we would get from applying the product rule. so the equation becomes:

$\left( \mu (x) y \right)' = \mu (x) f(x)$

$\Rightarrow \mu (x) y = \int \mu (x) f(x)~dx$

$\Rightarrow y = \frac { \int \mu (x) f(x) ~dx}{ \mu (x)}$ is our solution

remember that you will get an arbitrary constant from the integration, you have to divide this by $\mu (x)$ as well

**emily28** · July 12th 2007, 02:55 AM

after doing some reading on intergrating factors, from

ive used the intergrating factor, whereby f(x) = -4x and a(x) = 4

so g(x)= exp^(intergral of 4 dx)

so $g(x) = e^{4x}$

now using the formula y= 1/g(x).[intergral of f(x)g(x) dx]

so $y=1/e^{4x} [intergral -4xe^{4x}]<br />$

is this correct and how would i carry on from this?

**Jhevon** · July 12th 2007, 08:35 AM

Originally Posted by emily28

after doing some reading on intergrating factors, from

ive used the intergrating factor, whereby f(x) = -4x and a(x) = 4

so g(x)= exp^(intergral of 4 dx)

so $g(x) = e^{4x}$

now using the formula y= 1/g(x).[intergral of f(x)g(x) dx]

so $y=1/e^{4x} [intergral -4xe^{4x}]<br />$

is this correct and how would i carry on from this?

yes, that is fine, and incidentally, that is exactly the process i described to you above. i tried to save you the trouble of trying to get it from you text book, sometimes those things are ... let's just say, I hate the text that i used for differential equations. the only thing i like about it is that it gives the answers for even as well as odd problems, which is rare for textbooks to do

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