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Given this differential equation

$x^2y^3+y+(x^3y^2-x)y'=0$

I have to find an integrating factor, such that the equation becomes an exact differential equation.

I am pretty sure that the integrating factor must be $\mu (xy)$, so let $xy=z$.

After some algebra I got this differential equation:

$\mu '(z)(z+1)(y-x)=-2\mu (z)$

I have checked my calculations many times and I am certain, that this equation is correct.

However, I don't know what to do now, since I have my $\mu$ depending on the product $xy$.

How do I get $\mu (z)$? Or am I going in the wrong direction with this attempt?

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  • $\begingroup$ Are you sure the factor multiplying $y'$ is "$(x^3y^2 - x)$" and not "$(x^3y^2 + x)$"? The problem would simplify in the latter case (or, are you sure the "$x^2y^3 + y$" is correct, or is it "$x^2y^3-y$"?) $\endgroup$
    – Michael
    Nov 16, 2015 at 22:36
  • $\begingroup$ @Michael: Yes, I checked it and it is correct the way it's written. $\endgroup$
    – de_dust
    Nov 16, 2015 at 22:38
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    $\begingroup$ An integrating factor $\mu(x,y) = 1/(xy)$ seems to help. $\endgroup$
    – Michael
    Nov 16, 2015 at 22:51

1 Answer 1

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So, you can try multiplying both sides of the equation by $1/(xy)$.

Overall, your method using $\mu(z)$ (for $z=xy$) is correct, but you get a strange result $\mu'(z)(z+1)(y-x)=-2\mu(z)$, and I don't know how you got that. I get a very different answer when using your method. It is indeed a simple ODE for $\mu(z)$, and indeed leads to the same result as just directly multiplying by $1/(xy)$.


I got my answer of multiplying by $1/(xy)$ just by playing around. But you can do it more systematically your way:

Theory: Suppose $a(x,y) + b(x,y)y' = 0$. Multiply by $\lambda(x,y)$ to get: $$ \lambda(x,y)a(x,y) + \lambda(x,y)b(x,y)y'=0$$ We want a function $f(x,y)$ such that: \begin{align} &\partial f/\partial x = \lambda(x,y)a(x,y)\\ &\partial f/\partial y = \lambda(x,y)b(x,y) \end{align} If we find such, we can say $f(x,y(x))=C$. We should have $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$. So we want: $$ \frac{\partial [\lambda(x,y)a(x,y)]}{\partial y} = \frac{\partial [\lambda(x,y)b(x,y)]}{\partial x} $$ In this case you are told to try $\lambda(x,y)=\mu(z)$ for $z=xy$ (and $\partial z/\partial x$ and $\partial z/\partial y$ can easily be calculated) and so: $$ \frac{\partial [\mu(z)(x^2y^3 +y)]}{\partial y} = \frac{\partial [\mu(z)(x^3y^2-x)]}{\partial x} $$ Doing the derivatives and simplifying leads to a simple ODE in $\mu(z)$ (with no $x$ or $y$ variables).

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  • $\begingroup$ @de_dust : Were you able to fix the mistakes to get a proper ODE for $\mu(z)$ using your method? You end up within something like $h(z)\mu'(z) + r(z) =0$, with simple $h(z)$ and $r(z)$ functions. $\endgroup$
    – Michael
    Nov 16, 2015 at 23:12
  • $\begingroup$ Ok, I can't seem to find the integrating factor by myself. I get $\mu '(z)(x^2y^3+y-x^3y^2+x)=\mu (3x^2y^2-1)-\mu (z)(3x^2y^2+1)$ and after simplifying the term I end up with my equation. $\endgroup$
    – de_dust
    Nov 17, 2015 at 0:18
  • $\begingroup$ Could you please explain how you got your solution? $\endgroup$
    – de_dust
    Nov 17, 2015 at 0:18
  • $\begingroup$ What are you doing to get your equality? I will modify my answer to give more details. $\endgroup$
    – Michael
    Nov 17, 2015 at 6:23
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    $\begingroup$ Note that $d \mu/dx =(d\mu/d z)(dz/dx)$. $\endgroup$
    – Michael
    Nov 17, 2015 at 7:21

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