DE Exactness: (2x+y)dx-(x+6y)dy=0?

  • Thread starter Thread starter Eastonc2
  • Start date Start date
Eastonc2
Messages
19
Reaction score
0

Homework Statement


determine whether the DE, (2x+y)dx-(x+6y)dy=0, is exact

Homework Equations


i understand how to determine if they are exact, I just don't understand this specific instance. for my case, M(x,y)=2x+y, but would N(x,y)=(x+6y), or (-x-6y)?

The Attempt at a Solution


using my first N(x,y), the equation is exact, however, using the second, they are not exact.

Just need clarification at this point
 
Last edited:
Physics news on Phys.org
Hi Eastonc2! :smile:

Your N(x,y)=(-x-6y).
That is how it matches the definition of an exact DE.

But how did you determine that it was exact with the first N(x,y)?
Because I don't think it is.
 
ah, my fault, M(x,y)=2x+y

i was looking at the equation above it for that first part. the second part is correct though.
 
Ah, now I see your dilemma.

To make sure, perhaps you should try to find a function of which the partial derivatives match with M(x,y) and N(x,y).
Can you find such a function?
 
Have you studied the case where M and N are homogeneous of the same degree, as these are?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top