Question

Help with exact differential equations, see problem below....

Answer 1

Consider the equation \[(5x ^{2}y+6x ^{3}y ^{2}+4xy ^{2})dx+(2x ^{3}+3x ^{4}y+3x ^{2}y)dy=0\]
a. Show that the equation is not exact.

b. Multiply the equation by \[x ^{n}y ^{m}\]and determine values for n and m that make the resulting equation exact.

Answer 2

I know how to do part a. But I am having trouble with part b.

Answer 3

What is the definition for an exact differential equation? Perhaps there is a specific property about them that will allow you to find the values for m and n.

Answer 4

This might help out as well: http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx
See after the 1st example.

Answer 5

For a differential equation to be exact if \[\frac{ dM }{ dy }(x,y) =\frac{ dN }{ dx }(x,y)\]

Answer 6

so for this one I got \[\frac{ dM }{ dy }=5x ^{2}+12x^{3}y+8xy\]
\[\frac{ dN }{ dx }=6x ^{2}+12x ^{3}y+6xy\]

Answer 7

\[\frac{ dM }{ dy }\neq \frac{ dN }{ dx }\]
therefore it is not exact

Answer 8

Right. So then we want to multiply our original equation by \(x^n y^m\) and try to set them equal again.
This time, we might be able to find values of m and n that make that equation true.

Answer 9

\( x^n y^m \left[ (5x ^{2}y+6x ^{3}y ^{2}+4xy ^{2})dx+(2x ^{3}+3x ^{4}y+3x ^{2}y)dy=0\right] \)
\( (5x^{n+2}y^{m+1} + 6x^{n+3} y^{m+2} + 4x^{n+1} y^{m+2} ) \ dx \\ \qquad \qquad+ (2x^{n+3} + 3x^{n+4} y^{m+1} + 3x^{n+2}y^{m+1} ) \ dy = 0\)

Answer 10

\((5x^{n+2}y^{m+1} + 6x^{n+3} y^{m+2} + 4x^{n+1} y^{m+2} ) \ dx \\ \qquad \qquad+ (2x^{n+3}\color{red}{y^m} + 3x^{n+4} y^{m+1} + 3x^{n+2}y^{m+1} ) \ dy = 0\)
Missed that part.

So for example, we calculate the partial of M with respect to y again:
\(\dfrac{\partial M}{\partial y} = \color{blue}{5(m+1}) x^{n+2} y^m + \color{blue}{6(m+2)}x^{n+3} y^{m+1} + \color{blue}{4(m+2)} x^{n+1}y^{m+1} \)
Those coefficients are of special interest because both sides will have like-terms on each side, so their coefficents will be equal. Does that make sense?

Answer 11

yes, so,\[\frac{ dN }{ dx}=2(n+3)x ^{n+2}y ^{m}+3(n+4)x ^{n+3}y ^{m+1}+3(n+2)x ^{n+1}y ^{m+1}\]

Answer 12

Yep, that looks good.
So now \( \partial M / \partial y = \partial N / \partial x \).
With these equal, that means the coefficients have to be equal.
So the terms with x^(n+2) y^m: \(\color{blue}{5(m+1)} x^{n+2} y^m = \color{blue}{2(n+3)} x^{n+2} y^m \). So \( 5(m+1) = 2(n+3).\)

We can set up three equations with the coefficients here. One between terms with x^(n+3) y^(m+1), and one between the terms with x^(n+1) y^(m+1). Them we solve like a system of linear equations.

Answer 13

ok I have the equations set up like this

5m-2n=0
6m-3n=0
4m-3n=-2

Answer 14

When I solve for m and n I'm not getting the correct solutions

Answer 15

5m + 5 = 2n + 6 That doesn't become 5m - 2n = 0. 2*3 =/= 2+3. :)

Answer 16

wow!!! simple multiplication gets me every time!!!

Answer 17

ok so now i've got

5m-2n=1

6m-3n=0

4m-3n=-2

Answer 18

hold on i've got this.....found another simple mistake!

Answer 19

Yep, you can do it! :D

Answer 20

ok maybe i can't do it...i keep coming up with different answers

Answer 21

What answers were you getting?

Answer 22

Holy cow!!!!!!! I think I got it....

m=1
n=2

Answer 23

That was what I was getting as well. :)

Answer 24

I made a series of very simple but stupid multiplication and sign errors. I've been working on this for 5 hours!

Answer 25

Oftentimes the arithmetic/algebra becomes the hardest part of higher maths. :P

And when I just plugged in m=1 and n=2, the coefficients all do seem to be equal!

Answer 26

multiplying it through and taking the partials I got:
\[\frac{ dM }{ dy }=10x ^{4}y+18x ^{5}y ^{2}12x ^{3}y ^{2}\]
\[\frac{ dN }{ dx}=10x ^{4}y+18x ^{5}y ^{2}12x ^{3}y ^{2}\]

YAY!!!!!!

Answer 27

Thanks for your help!!!!!

Answer 28

You're welcome! :D