Question

Question about Exact Equations

Answer 1

I don't know how to begin

Answer 2

shouldnt the integrating factor be \[e^{-\int f(y)dy}\] or something?

Answer 3

dude u kno wat an xact equation is?

Answer 4

i do

Answer 5

so if u multipy the integrating factor......resultant equation is xact.........i think so.....give it a try.......

Answer 6

are you suppose to prove it algebraically?

Answer 7

ok ill try that

Answer 8

\[M+Ny'=0\]\[M\text dx+N\text dy=0\]\[MQ(y)\text dx+NQ(y)\text dy=0\]the equation is exact if\[\frac{\partial MQ(y)}{\partial y}=\frac{\partial NQ(y)}{\partial x}\]

Answer 9

\[Q(y)\frac{\partial M}{\partial y}+MQ'(y)=Q(y)N'\]

Answer 10

Ah that seems a bit more easy!!
First order Linear ODE

Answer 11

wait did u multiply q(x)......u shud multiply r(x)

Answer 12

\[ M \frac{\partial Q(y)}{\partial y} + Q(y)(M_y - N_x)=0 \]

Answer 13

\[M+Ny'=0\]\[M\text dx+N\text dy=0\]\[MR(y)\text dx+NR(y)=0\]\[\frac{\partial MR(y)}{\partial y}=\frac{\partial NR(y)\text dx}{\partial x}\]\[\frac{\partial MR(y)}{\partial y}=R(y)\frac{\partial N}{\partial x}\]\[M\frac{R(y)}{\partial y}+R(y)M_y=R(y)N_x\]\[M\frac{R(y)}{\partial y}+R(y)\left(M_y-N_x\right)=0\]

Answer 14

\[\frac{R(y)}{\partial y}+\frac{R(y)}{M}\left(M_y-N_x\right)=0\]

Answer 15

i think i know what to do now

Answer 16

That's fine ...
That's first order linear ODE in R(y)
it's value is
\[ C_1 e^{\int \frac{(M_y−N_x)}{M} dy }\]

Where C1 is a constant ...
if you haven't done this type of differential equation then check this video
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Answer 17

Also since it's IF ... ignore C_1

Answer 18

Did we make some mistake ??? in + -

also check it here
http://www.sosmath.com/diffeq/first/lineareq/lineareq.html

Answer 19

i have watched that video 5 times already lolz

Answer 20

Ah ... it's allright!! I made mistake ...
\[ C_1 e^{-\int \frac{(M_y−N_x)}{M} dy }\]

Answer 21

did i get Q and R mixed up again?

Answer 22

Whatever ...
it doesn't make any difference ... gotta go
catch you later!!

Answer 23

thanks for your help

Answer 24

yw

Answer 25

\[M+Ny'=0\]\[M\text dx+N\text dy=0\]\[MQ(y)\text dx+NQ(y)=0\]\[\frac{\partial MQ(y)}{\partial y}=\frac{\partial NQ(y)}{\partial x}\]\[\frac{\partial MQ(y)}{\partial y}=Q(y)\frac{\partial N}{\partial x}\]\[M\frac{Q(y)}{\partial y}+Q(y)M_y=Q(y)N_x\]\[M\frac{Q(y)}{\partial y}+Q(y)\left(M_y-N_x\right)=0\]\[\frac{Q(y)}{\partial y}+\frac{Q(y)}{M}\left(M_y-N_x\right)=0\]\[Q(y)'+Q(y)\frac{\left(M_y-N_x\right)}{M}=0\]\[R(y)=e^{\int\limits^y Q(t)\text dt}\]\[\left(Q(y)R(y)\right)'=0\]\[\frac{\partial Q(y)R(y)}{\partial y}=0\]

Answer 26

\[Q(y)R(y)=y+c_1\]

Answer 27

i have gotten mixed up with the Q and R.

Answer 28

can someone see where i am getting mixed up ?

Answer 29

and i dont really know how to tell if i have answered the question

Answer 30

I think the proof should last only upto the finding IF .. you have already assumed IF has only y and and multiplying by IF makes it exact DE

Answer 31

\[M+Ny'=0\]\[M\text dx+N\text dy=0\]\[MR(y)\text dx+NR(y)\text dy=0\]\[\frac{\partial MR(y)}{\partial y}=\frac{\partial NR(y)}{\partial x}\]\[\frac{\partial MR(y)}{\partial y}=R(y)\frac{\partial N}{\partial x}\]\[M\frac{\partial R(y)}{\partial y}+R(y)M_y=R(y)N_x\]\[M\frac{\partial R(y)}{\partial y}+R(y)\left(M_y-N_x\right)=0\]\[\frac{\partial R(y)}{\partial y}+\frac{R(y)}{M}\left(M_y-N_x\right)=0\]\[R(y)'+R(y)\frac{\left(M_y-N_x\right)}{M}=0\]\[R(y)=e^{\int\limits^y\frac{\left(M_y-N_x\right)}{M}\text dy}=e^{\int\limits^y Q(t)\text dt}\]
I sean to have dropped a minus sign somewhere
my \(Q(t)\) should be \(-Q(t)\)