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

1. ## [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.

2. 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 $dv = e^{-x}$...

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

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

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

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

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

$x + z = Ce^{x}$

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

3. 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..

4. Are you sure it's not $(-x + z)e^{-x} = C$?

5. 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..

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

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