# Thread: [SOLVED] differential equation given integrating factor

1. ## [SOLVED] differential equation given integrating factor

Show that given function μ is an integrating factor and solve the differential equation..

y^2 dx + (1 + xy) dy = 0 ; μ(x) = e^xy

let M = y^2
N = (1 + xy)

dM/dy = 2y dN/dx = y hence, not exact equation.

times μ(x) = e^xy to the not exact equations...

2y(e^xy) dx + y(e^xy) dy = 0

let M = 2y(e^xy)
N = y(e^xy)

dM/dy = 2(e^xy) + 2y(e^y) ---> apply product rule

dN/dx = 0(e^xy) + y(e^y) ---> apply product rule

the problem is.. the equations still not the exact equations..
How to proceed?

2. Originally Posted by nameck
Show that given function μ is an integrating factor and solve the differential equation..

y^2 dx + (1 + xy) dy = 0 ; μ(x) = e^xy

let M = y^2
N = (1 + xy)

dM/dy = 2y dN/dx = y hence, not exact equation.

times μ(x) = e^xy to the not exact equations...

2y(e^xy) dx + y(e^xy) dy = 0

let M = 2y(e^xy)
N = y(e^xy)

dM/dy = 2(e^xy) + 2y(e^y) ---> apply product rule

dN/dx = 0(e^xy) + y(e^y) ---> apply product rule

the problem is.. the equations still not the exact equations..
How to proceed?
$y^2\,dx + (1 + xy)\,dy = 0$

$y^2\,dx = -(1 + xy)\,dy$

$y^2\,\frac{dx}{dy} = -(1 + xy)$

$y^2\,\frac{dx}{dy} = -1 - xy$

$\frac{dx}{dy} = -y^{-2} - xy^{-1}$

$\frac{dx}{dy} + xy^{-1} = -y^{-2}$.

Now this is a linear DE in $x$.

So the integrating factor is $e^{\int{y^{-1}\,dy}} = e^{\ln{y}} = y$

Multiplying through by the integrating factor gives:

$y\,\frac{dx}{dy} + x = -y^{-1}$

$\frac{d}{dy}(xy) = -y^{-1}$

$xy = \int{-y^{-1}\,dy}$

$xy = -\ln{|y|} + C$

$x = -y^{-1}\ln{|y|} + Cy^{-1}$.

3. got it.. so... the integrating factor given, is not the correct integrating factor... right?

4. Originally Posted by nameck
got it.. so... the integrating factor given, is not the correct integrating factor... right?
Unless there's a way to make it a linear equation in $y$. However, I couldn't find one, and since you can't rewrite the solution as $y$ in terms of $x$, I doubt that you can...

5. thanks Prove It!!