# Problem with Integrating Factor

• Jan 24th 2011, 02:52 AM
StaryNight
Problem with Integrating Factor
I'm trying to solve the differential equation dy/dx + (2-3x^2)(x^-3)y = 1. I've integrated (2-3x^2)(x^-3) and taken the exponential of this to give an integrating factor of (x^-3)e^(x^2). However, when I differentiate the integrating factor multiplied by y, I don't end up with the left hand side of the equation. I must have gone wrong somewhere in my working, however, I've checked it several times!
• Jan 24th 2011, 03:04 AM
mr fantastic
Quote:

Originally Posted by StaryNight
I'm trying to solve the differential equation dy/dx + (2-3x^2)(x^-3)y = 1. I've integrated (2-3x^2)(x^-3) and taken the exponential of this to give an integrating factor of (x^-3)e^(x^2). However, when I differentiate the integrating factor multiplied by y, I don't end up with the left hand side of the equation. I must have gone wrong somewhere in my working, however, I've checked it several times!

I think you'll find that in the integrating factor it's $\displaystyle e^{-\frac{1}{x^2}$ not $\displaystyle e^{x^2}$. Note: $\displaystyle -\frac{1}{x^2} \neq x^2$ .... it seems you're confusing $\displaystyle -\frac{1}{x^2}$ with $\displaystyle \frac{1}{x^{-2}}$ ....
• Jan 24th 2011, 03:07 AM
alexmahone
Quote:

Originally Posted by StaryNight
I'm trying to solve the differential equation dy/dx + (2-3x^2)(x^-3)y = 1. I've integrated (2-3x^2)(x^-3) and taken the exponential of this to give an integrating factor of (x^-3)e^(x^2). However, when I differentiate the integrating factor multiplied by y, I don't end up with the left hand side of the equation. I must have gone wrong somewhere in my working, however, I've checked it several times!

$\displaystyle \displaystyle \int \frac{2-3x^2}{x^3}=-\frac{1}{x^2}-3lnx+C$

Integrating Factor = $\displaystyle \displaystyle e^{-\frac{1}{x^2}-3lnx}=e^{-\frac{1}{x^2}}e^{-3lnx}=x^{-3}e^{-\frac{1}{x^2}}$
• Jan 24th 2011, 03:20 AM
FernandoRevilla
We can rewrite the equation in the form:

$\displaystyle Pdx+Qdy=0\;,\quad P=(1-3x^2)y\;,\;Q=x^3$

and:

$\displaystyle \dfrac{1}{Q}\left(\dfrac{{\partial P}}{{\partial y}}-\dfrac{{\partial Q}}{{\partial x}}\right)=\ldots=\dfrac{1}{x^3}-\dfrac{6}{x}=F(x)$

so, an integrating factor is:

$\displaystyle \mu(x)=e^{\int F(x)\;dx}$

Fernando Revilla

Edited: Sorry, I misread the OP's equation.