Question

\[(2t^3y-y^3)\text dt-2t^4\text dy=0\]

Answer 1

\[\frac{\partial M}{\partial y}=2t^3-3y^2\]\[\frac{\partial N}{\partial t}=-8t^3\]\[\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial t}\]

Answer 2

\[R=R(t,y)=t^my^m\]\[\frac{\partial MR(t,y)}{\partial y}=\frac{\partial NR(t,y)}{\partial t}\]\[R\frac{\partial M}{\partial y}+R_yM=R\frac{\partial N}{\partial t}+R_tN\]

Answer 3

\[R\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial t}\right)=R_tN-R_yM\]
\[t^my^n\left(2t^3y-3y^2+8t^3\right)=mt^{m-1}y^n(2t^3y−y^3)+nt^my^{n-1}(2t^4)\]

Answer 4

*\[R=R(t,y)=t^my^n\]

Answer 5

how to find \(m,n\)

Answer 6

I haven't tried this yet, but you're trying to solve a exact equation, mind if I ask you what you did to make this exact? There is this annoying negative sign

Answer 7

i assumed an intergrating facor was of the form \( R(t,y)=t^my^n\)

Answer 8

which negative sign do you mean?

Answer 9

It seems the method of homogenous equations is an easier method of solving this problem.

Answer 10

well im trying to do it this way

Answer 11

\[M(x.y)+N(x,y)\]

hmm yes I understand that part, what bothers me is that the function above doesn't look like a characteristcal exact equation to me. Sorry it's a pain to LaTeX here at the moment, I am on my DOs machine, not used to it.

Answer 12

If I recall it correctly, the plus in between is essential to be exact.

Answer 13

N=-N
?

Answer 14

Ok I solved it. Only thing is I didn't use an integrating factor unfortunately. Lemme try to see if I can.

Answer 15

I will have to work through it myself at first too, sorry if I caused more questions then answers, but that negative sign really seems alien to me at first glance, various substitution will sure get rid of it, I just can't recall if that influences the result or not.

Answer 16

am i going about this the right way?

Answer 17

To be honest, it doesn't seem like it. The most feasible thing to find m and n in this case is to solve\[R(dM/dy - dN/dt) = 0\] but it doesn't look like you can get a solution by hand. Since this is a homogenous equation, I suggest you tackle that direction since it's a lot more straightforward.

Answer 18

\[R\frac{\partial M}{\partial y}+R_yM=R\frac{\partial N}{\partial t}+R_tN\]
\[t^my^n(2t^3)+nt^my^{n-1}(2t^3y)=t^my^n(-8t^3)+mt^{m-1}y^n(-2t^4)\]
\[2t^{m+3}y^n+2nt^{m+3}y^{n}=-8t^{m+3}y^n-2mt^{m+3}y^n\]
\[(10+2n+2m)t^{m+3}y^n=0\]
\[(10+2n+2m)=0\]

Answer 19

Ahh yes, sorry I meant to say solve for \[d/dy(RM) = d/dx(RN)\]

Answer 20

im not sure what your saying @dominusscholae

Answer 21

^basically what you did in the last post. You seem to be doing the correct thing in order to find m and n. So (at least to me) any combination of m and n that satisfies 10 +2m + 2n = 0. Thing is, when I plug any value satisfying this, I'm not getting an exact equation for some reason.

Answer 22

i need two equations to solve two variables, how do i get the other/