differential equations - Page 2

topsquark · #16 $Old$ July 11th, 2007, 09:17 AM

Quote:

Originally Posted by emily28 $View Post$

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 · #17 $Old$ July 11th, 2007, 10:16 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 · #18 $Old$ July 11th, 2007, 01:34 PM

does anyone have any suggestions/

Krizalid · #19 $Old$ July 11th, 2007, 01:38 PM

Quote:

Originally Posted by emily28 $View Post$

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

Find an integrating factor defined by

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

topsquark · #20 $Old$ July 11th, 2007, 02:30 PM

Quote:

Originally Posted by emily28 $View Post$

does anyone have any suggestions/

See here.

-Dan

Jhevon · #21 $Old$ July 11th, 2007, 09:00 PM

Quote:

Originally Posted by emily28 $View Post$

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 · #22 $Old$ July 12th, 2007, 04:55 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 $y=1/e^{4x} [intergral -4xe^{4x}]$

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

Jhevon · #23 $Old$ July 12th, 2007, 10:35 AM

Quote:

Originally Posted by emily28 $View Post$

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

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