Shouldn't you remove the integrating factor?

• Jan 1st 2011, 04:35 PM
MSUMathStdnt
Shouldn't you remove the integrating factor?
One method for doing an ODE is to try to make it an exact ODE by multiplying by an integrating factor. Why don't you have to remove the integrating factor afterwards?

For instance, in this example, the problem was not an exact DiffEQ:
$\left( 3xy+y^2 \right) + \left( x^2+xy \right) y' = 0$

$M(x,y)=3xy+y^2 \hspace{20 mm} N(x,y)=x^2+xy$
$M_y=3x+2y \hspace{30 mm} N_x=2x+y$

$M_y \ne N_x$, so NOT an exact DiffEQ.

Next is to look for an integrating factor, which we call $\mu(x)$:
$\mu(x) \left( 3xy+y \right) + \mu(x) \left( x^2+xy \right) y' = 0$

$M(x,y)=\mu(x)\left( 3xy+y^2 \right) \hspace{20 mm} N(x,y)=\mu(x)\left( x^2+xy \right)$
$M_y=\mu(x)(3x+2y) \hspace{30 mm}= N_x=\mu'(x)(x^2+xy)+\mu(x)(2x+y)$
$\mu(x)(x+y)=\mu'(x)(x^2+xy)$
$\mu(x)=\mu'(x)x=\fracd{d\mu}{dx}x$
$\frac{dx}{x}=\frac{d\mu}{\mu} \implies \mu(x)=x$

So, back to the top, multiply the whole equation by x:
$\left( 3x^2y+xy^2 \right) + \left( x^3+x^2y \right) y' = 0$

$\int(3x^2y+xy^2)dx = x^3y+\frac12x^2y^2+f(y)$
$\int(x^3+x^2y)dy = x^3y+\frac12x^2y^2+g(x)$
$f(x)=g(x)=C$(some constant)

$\psi(x,y)=x^3y+\frac12x^2y^2=C$

Why don't we have to compensate for $\mu(x)=x$ after this last step? Notice the $\psi_x$ and $\psi_y$ do not equal the starting $M$ and $N$:

$\psi_x=3x^2y+xy^2 \hspace{30 mm} \psi_y=x^3+x^2y$

And dividing psi by x also doesn't work (although dividing the partial derivatives of psi by x works). What gives here?

Thanks (sorry for the too long post),
• Jan 1st 2011, 04:42 PM
dwsmith
$\left( 3x^2y+xy^2 \right) + \left( x^3+x^2y \right) y' = 0\Rightarrow x(3xy+y^2)dx=-x(x^2+xy)dy$

$\displaystyle (3xy+y^2)dx=-(x^2+x+y)dy \ \mbox{multiple LHS by 1}$

$\displaystyle\Rightarrow \frac{x}{x}(3xy+y^2)dx=-(x^2+xy)dy\Rightarrow x(3xy+y^2)dx=-x(x^2+xy)dy$

You don't need to account for it since it was a multiplication by 1.
• Jan 1st 2011, 06:55 PM
MSUMathStdnt
OK. But I'm still lost.
What's the answer, then. Is psi(x,y) actually y(x,y) or something?
• Jan 1st 2011, 07:15 PM
dwsmith
Quote:

Originally Posted by MSUMathStdnt
$\left( 3x^2y+xy^2 \right) + \left( x^3+x^2y \right) y' = 0$

$\int(3x^2y+xy^2)dx = x^3y+\frac12x^2y^2+f(y)$

$\displaystyle \frac{\partial f}{\partial y}=f_y=x^3+x^2y$

$\displaystyle f_y=x^3+x^2y=N(x,y)=x^3+x^2y\Rightarrow f_y=0\Rightarrow \int f_ydy=\int 0dy\Rightarrow f(y)=c$

$\displaystyle g(x,y)=x^3y+\frac{x^2}{2}y^2+c$

My original post here was correct. I don't know I changed it, sorry.