# a general Differential Equations question

• Nov 4th 2012, 11:57 AM
smacker
a general Differential Equations question
Hello all ,
i have the equation
Code:

`(1+x^2)y'+y=arctgx`
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 !!
• Nov 4th 2012, 01:56 PM
topsquark
Re: a general Differential Equations question
Quote:

Originally Posted by smacker
Hello all ,
i have the equation
Code:

`(1+x^2)y'+y=arctgx`
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
• Nov 4th 2012, 02:04 PM
smacker
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 :
Quote:

lnx\x
and i want to call t=lnx
so it will come out tdt ?
the 1/x becomes the dt ?
if so math is awsome