# [SOLVED] find integrating factor and solve the equation 2

• March 27th 2010, 07:52 PM
nameck
[SOLVED] find integrating factor and solve the equation 2
Question:

(1 - x - z)dx + dz = 0

let M = (1 - x - z) ; N = 1

dM/dz = -x ; dN/dx = 0
hence, not exact.

integrating factor...

1/N[ dM/dz - dN/dx ] = 1[-x]
.................................= -x

e^ ∫-x dx = e ^ [(-x^2)/2]

times integrating factor with DE..

{e ^ [(-x^2)/2] -x e ^ [(-x^2)/2] - z e ^ [(-x^2)/2]}dx +
e ^ [(-x^2)/2]dz = 0

dM/dz = -e ^ [(-x^2)/2] ; dN/dx = -xe ^ [(-x^2)/2]

the problem is, i calculated the integrating factor.. but, the equations still not the exact calculation.
• March 27th 2010, 08:18 PM
Prove It
Quote:

Originally Posted by nameck
Question:

(1 - x - z)dx + dz = 0

let M = (1 - x - z) ; N = 1

dM/dz = -x ; dN/dx = 0
hence, not exact.

integrating factor...

1/N[ dM/dz - dN/dx ] = 1[-x]
.................................= -x

e^ ∫-x dx = e ^ [(-x^2)/2]

times integrating factor with DE..

{e ^ [(-x^2)/2] -x e ^ [(-x^2)/2] - z e ^ [(-x^2)/2]}dx +
e ^ [(-x^2)/2]dz = 0

dM/dz = -e ^ [(-x^2)/2] ; dN/dx = -xe ^ [(-x^2)/2]

the problem is, i calculated the integrating factor.. but, the equations still not the exact calculation.

$(1 - x - z)\,dx + dz = 0$

$1 - x - z + \frac{dz}{dx} = 0$

$\frac{dz}{dx} - z = x - 1$.

Now the integrating factor is $e^{-1\,dx} = e^{-x}$.

Multiplying through by the integrating factor gives:

$e^{-x}\,\frac{dz}{dx} - e^{-x}z = e^{-x}(x - 1)$

$\frac{d}{dx}(e^{-x}z) = e^{-x}(x - 1)$

$e^{-x}z = \int{e^{-x}(x - 1)\,dx}$.

Now using integration by parts with $u = x - 1$ and $dv = e^{-x}$...

$e^{-x}z = -e^{-x}(x - 1) - \int{-e^{-x}\,dx}$

$e^{-x}z = -e^{-x}(x - 1) - e^{-x} + C$

$z = -(x - 1) - 1 + Ce^x$

$z = -x + 1 - 1 + Ce^{x}$

$z = -x + Ce^{x}$

$x + z = Ce^{x}$

$(x + z)e^{-x} = C$.
• March 27th 2010, 08:42 PM
nameck
Prove It....

i kinda stuck here.. the answer given is (x + z)e^-x = C
which is not same as the answer u calculated above.. (Worried)
• March 27th 2010, 08:47 PM
Prove It
Are you sure it's not $(-x + z)e^{-x} = C$?
• March 27th 2010, 08:54 PM
nameck
got it!!
from here...
[tex]

$e^{-x}z = -e^{-x}(x - 1) - e^{-x} + C$

but.. the way u solve this equation is easier than what i've been told in my lecture.. i need to integrate, then differentiate, the compare the value of constant, C.. which is more confusing and the chance for me to make mistakes is high..
• March 27th 2010, 08:58 PM
Prove It
Quote:

Originally Posted by nameck
got it!!
from here...
[tex]

$e^{-x}z = -e^{-x}(x - 1) - e^{-x} + C$