a general Differential Equations question

Hello all ,

i have the equation

now i managed to find the formula for it and got to:

Code:

` `

y=e^-arctanx[~(arctanx*e^tanx*dx)/(1+x^2)]

where ~ is the sign for integral

now i'm supposed to do t=arctanX

but don't understand where i get the dt from ?

how do i find the DT is what i'm asking here :p

thanks to all the helpers !!

Re: a general Differential Equations question

Quote:

Originally Posted by

**smacker** Hello all ,

i have the equation

now i managed to find the formula for it and got to:

Code:

` `

y=e^-arctanx[~(arctanx*e^tanx*dx)/(1+x^2)]

where ~ is the sign for integral

now i'm supposed to do t=arctanX

but don't understand where i get the dt from ?

how do i find the DT is what i'm asking here :p

thanks to all the helpers !!

It looks like a standard substitution.

$\displaystyle t = arctan(x)$

What is the derivative of arctan(x)? Then compare that form with your integrand. (You'll have to find dx in terms of dt so as a hint you might want to do this by x = tan(t) etc.)

-Dan

Re: a general Differential Equations question

well the answer will be 1/1+x^2

but the question is - is it always the derrative ?

like for example :

and i want to call t=lnx

so it will come out tdt ?

the 1/x becomes the dt ?

if so math is awsome