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fishfood12m's question at Yahoo! Answers regarding special integrating factors

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  • #1


Staff member
Feb 24, 2012
Here is the question:

Integrating Factor Help Please!?

I am preparing for a test and need to be able to find the integrating factor of problems similar to:

ydx + (2xy-e^(-2y))dy=0

All I want is how to find the integrating factor. From what I learned u'=U*(My - Nx) / N

That doesn't seem to work for me. Perhaps you can explain how to find it with step-by-step detail.

Here is a link to the question:

Integrating Factor Help Please!? - Yahoo! Answers

I have posted a link there so the OP can find my response.
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Feb 24, 2012
Hello fishfood12m,

We are given to solve:

$\displaystyle y\,dx + (2xy-e^{-2y})\,dy=0$

We should first verify that it is not exact:

$\displaystyle \frac{\delta}{\delta y}(y)=1$

$\displaystyle \frac{\delta}{\delta x}(2xy-e^{-2y})=2y$

Since the two partials are not equal, the equation is not exact.

Next, we look at the expression:

$\displaystyle \frac{\frac{\delta}{\delta y}(y)-\frac{\delta}{\delta x}(2xy-e^{-2y})}{2xy-e^{-2y}}=\frac{1-2y}{2xy-e^{-2y}}$

Since this does not depend on $x$ alone, we next look at:

$\displaystyle \frac{\frac{\delta}{\delta x}(2xy-e^{-2y})-\frac{\delta}{\delta y}(y)}{y}=\frac{2y-1}{y}=2-\frac{1}{y}$

Since this depends only on $y$, we compute the integrating factor as follows:

$\displaystyle \mu(y)=e^{\int2-\frac{1}{y}\,dy}=e^{2y-\ln|y|}=\frac{e^{2y}}{y}$

We now multiply the ODE by $\mu(y)$ observing we are losing the trivial solution $y\equiv0$:

$\displaystyle \frac{e^{2y}}{y}\cdot y\,dx + \frac{e^{2y}}{y}(2xy-e^{-2y})\,dy=0$

$\displaystyle e^{2y}\,dx + \left(2xe^{2y}-\frac{1}{y} \right)\,dy=0$

We can now easily see that the equation is exact. Even though you asked only for the integrating factor, I will go ahead and solve the equation.

Since our equation is exact, we may state:

$\displaystyle \frac{\delta F}{\delta x}=e^{2y}$

Integrating with respect to $x$ we get:

$\displaystyle F(x,y)=\int e^{2y}\,dx+g(y)=xe^{2y}+g(y)$

Now, to determine $g(y)$, we will take the partial derivative with respect to $y$ of both sides of the above equation and substitute $\displaystyle 2xe^{2y}-\frac{1}{y}$ for $\displaystyle \frac{\delta F}{\delta y}$:

$\displaystyle 2xe^{2y}-\frac{1}{y}=2xe^{2y}+g'(y)$

$\displaystyle -\frac{1}{y}=g'(y)$

Hence, integrating, and choosing the constant of integration to be zero, we find:

$\displaystyle g(y)=-\ln|y|$

And so we have:

$\displaystyle F(x,y)=xe^{2y}-\ln|y|$

Thus, the solution to the ODE is given implicitly by:

$\displaystyle C=xe^{2y}-\ln|y|$

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
Another way (without using integrating factor): we can write the equation in the form $\dfrac{dx}{dy}+2x=\dfrac{e^{-2y}}{y}$ which is linear on the dependent variable $x=x(y)$. Now, and using a well-known theorem, the general solution of the equation $x'+p(y)x=q(y)$ is

$xe^{\int pdy}-\displaystyle\int qe^{\int pdy}\;dy=C$
In our case, $p=2,q=\dfrac{e^{-2y}}{y}$ so,
$xe^{2y}-\displaystyle\int \frac{e^{-2y}}{y}\;e^{2y}\;dy=C$


$xe^{2y}-\ln |y|=C$