Is there a systematic way to determine an integrating factor #mu(x,y)# of the form #x^n y^m#, given a not-necessarily-exact differential equation?

Question

My book covers special integrating factors #mu# that are functions of only #x# or only #y#, but kinda glosses over how to find an integrating factor that is a function of #x# AND #y#.

Example equation:
#(2y^2 - 6xy)dx + (3xy - 4x^2)dy = 0#

The integrating factor was #mu(x,y) = xy#, and the solution was #F(x,y) = x^2y^3 - 2x^3y^2 = C#.

I was able to figure out what the integrating factor was, and solve the equation, but I had to assume that #n = m#, which is not something I think I should need to do.

Elijah Abram · Accepted Answer

If you have: #M(x,y) dx + N(x,y)dy=0# And the equation is not an exact Differential Equation, ie #(partial M) /(partial y) != (partial N) / (partial x) # Then you must convert the equation into an exact differential equation by multiplying by an integrating factor #mu(x,y)# to get #mu(x,y)M(x,y) dx + mu(x,y)N(x,y)dy=0# Where  #(partial (muM)) /(partial y) = (partial (muN)) / (partial x) # That's all well and good but In order to find such an integrating factor #mu(x,y)# you can do some manipulation and eventually establish the need to solve the partial differential equation: # "     " M (partial mu) /(partial y)  - N(partial mu) /(partial x)  + ((partial M) /(partial y)  - (partial N) /(partial x) ) mu # = 0
or,
# "     " Mmu_y-Nmu_x + (M_y-N_x)mu=0# which in general is a harder problem to solve! If the given differential equation is "designed" to be solved (eg in an exam rather than a real life equation) then it will often be the case that: either
# "     " mu(x,y) = mu(x)#, a function of #x# alone
or
# "     " mu(x,y) = mu(y)#, a function of #y# alone In which case the above PDE can easily be solved to give: # "     " mu(y) = exp(int ((partial M) /(partial y)-(partial N) /(partial x))/M dy) = e^(int (M_y-N_x)/M dy)#
or
# "     " mu(x) = exp(int ((partial M) /(partial y)-(partial N) /(partial x))/N dx) = e^(int (M_y-N_x)/N dx)# respectively. But, in general finding the integrating factor will not be possible and so the Differential Equation would be solved numerically rather than finding an analytical solution.   In the real world, It is always possible to find a series solution but this approach is particularly cumbersome (and is often the approach used by a computer for a numerical solution)

Christopher Agresta · Answer

undefined Yes, there is a systematic way to determine an integrating factor $ \mu(x,y) $ of the form $ x^n y^m $ for a not-necessarily-exact differential equation. One common approach is to use the method of "integrating factors." Given a differential equation of the form $ M(x,y)dx + N(x,y)dy = 0 $, where $ M $ and $ N $ are functions of $ x $ and $ y $, you can follow these steps:

1. Check if the equation is exact by verifying if $ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $. If it is exact, proceed to solve it using standard methods. If not, continue to step 2.

2. Find an integrating factor $ \mu(x,y) $ of the form $ x^n y^m $ by solving the equation $ \frac{\frac{\partial(\mu M)}{\partial y} - \frac{\partial(\mu N)}{\partial x}}{\mu} = \text{some function of } x \text{ and } y $.

3. Set up the partial differential equation $ \frac{\partial(\mu M)}{\partial y} - \frac{\partial(\mu N)}{\partial x} = \text{some function of } x \text{ and } y $ and solve it to find $ \mu(x,y) $.

4. Once you have determined $ \mu(x,y) $, multiply both sides of the given differential equation by $ \mu(x,y) $ and check if the resulting equation is exact. If it is, proceed to solve it using standard methods.

Following these steps systematically can help you determine an integrating factor $ \mu(x,y) $ of the form $ x^n y^m $ for a not-necessarily-exact differential equation.