Integrating factor (ODE)

**matlabnoob** · December 27th 2009, 08:28 AM

the problem goes..

The integrating factor for the ODE dxdt - 2xt = t is the integrating factor e*(-2integral(tdt)) = e^-t^2

multiplying the ODE by the integrating factor:
ddt((e^-t^2)*x) = t(e^-t^2)

integrate : e^-t^2 = -0.5(e^-t^2) + c

how does one get the integation bit?? i cant understand it at all =( .. especially the -0.5 ( .... ) +c bit

i tried by parts..but =S.. i end up integrating again and again....and again =| wow

**galactus** · December 27th 2009, 09:43 AM

As you said, The integrating factor is $e^{-t^{2}}$

Multiply through:

$\frac{d}{dt}[xe^{-t^{2}}]=te^{-t^{2}}$

Integrating 'negates' the derivative on the left:

$xe^{-t^{2}}=\frac{-1}{2}e^{-t^{2}}+C$

Divide by $e^{-t^{2}}$ and remember the log laws:

$x=Ce^{t^{2}}-\frac{1}{2}$

**matlabnoob** · December 27th 2009, 10:38 AM

Originally Posted by galactus

As you said, The integrating factor is $e^{-t^{2}}$

Multiply through:

$\frac{d}{dt}[xe^{-t^{2}}]=te^{-t^{2}}$

Integrating 'negates' the derivative on the left:

$xe^{-t^{2}}=\frac{-1}{2}e^{-t^{2}}+C$

Divide by $e^{-t^{2}}$ and remember the log laws:

$x=Ce^{t^{2}}-\frac{1}{2}$

what do you mean by 'negates'? [sorry if its a stupid simple question!]
i can see how if i differentiate..i get the original equation.. but i cannt see how to integrate it yet! =(
if i do it by parts... i have to keep integrating forever...

do you just see the answer clearly by looking?

**Prove It** · December 27th 2009, 05:43 PM

"Undoes" is probably a better word than "negates".

Integrating undoes the derivative.

So if $\frac{d}{dt}[xe^{-t^2}] = te^{-t^2}$

$\int{\frac{d}{dt}[xe^{-t^2}]\,dt} = \int{te^{-t^2}\,dt}$

The left hand side is $xe^{-t^2} + C_1$ .

Now use a $u$ substitution on the RHS.

Let $u = -t^2$ so that $\frac{du}{dt} = -2t$ .

So the RHS becomes

$\int{te^{-t^2}\,dt} = -\frac{1}{2}\int{-2te^{-t^2}\,dt}$

$= -\frac{1}{2}\int{e^u\,\frac{du}{dt}\,dt}$

$= -\frac{1}{2}\int{e^u\,du}$

$= -\frac{1}{2}e^u + C_2$

$= -\frac{1}{2}e^{-t^2} + C_2$ .

Therefore, upon integrating both sides, the DE is

$xe^{-t^2} + C_1 = -\frac{1}{2}e^{-t^2} + C_2$

$xe^{-t^2} = -\frac{1}{2}e^{-t^2} + C$ , where $C = C_2 - C_1$

$x = -\frac{1}{2} + Ce^{t^2}$ .

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