Question

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

Answer 1

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

Answer 2

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

Answer 3

\[t^my^n(2t^3)+nt^my^{n-1}(2t^3y)=t^my^n(-8t^3)+mt^{m-1}y^n(-2t^4)\]
\[2t^{m+3}y^n+2nt^{m+3}y^{n}=-8t^{m+3}y^n-2mt^{m+3}y^n\]
\[(10+2n+2m)t^{m+3}y^n=0\]

Answer 4

Can you tell me what topic is this ??

Answer 5

first order differential equations - exact equations - exact equations with integrating factor of both variables

Answer 6

i think i've worked it out now

Answer 7

its a differnt way then i was expecting

Answer 8

But you are getting the same result..

For this topic I have to google and then I can understand what part Mr Uncle Rocks is doing..


I am unfamiliar with it..

Answer 9

i wasent getting any result the first time i was getting stuck..

Answer 10

Suppose I want to search this topic then can you tell me what should I write their so that I can find how to solve these types of problems?

Any particular name to search this topic ..

Answer 11

http://www.google.com.au/search?sugexp=chrome,mod=4&sourceid=chrome&ie=UTF-8&q=first+order+differential+equations+-+exact+equations+-+exact+equations+with+integrating+factor+of+both+variables

Answer 12

\[\frac{\partial MR(t,y)}{\partial y}=\frac{\partial NR(t,y)}{\partial t}\]
\[\frac{\partial }{\partial y}\left((2t^3y-y^3)t^my^n\right)=\frac{\partial }{\partial t}\left((-2t^4)t^my^n\right)\]\[\frac{\partial }{\partial y}\left(2t^{m+3}y^{n+1}-t^my^{n+3}\right)=\frac{\partial }{\partial t}\left(-2t^{m+4}y^n\right)\]\[2(n+1)t^{m+3}y^{n}-(n+3)t^my^{n+2}=-2(m+4)t^{m+3}y^n\]
\[2(n+1)=-2(m+4);\qquad\qquad (n+3)=0\]\[n+1=-m-4;\qquad\qquad n=-3\]\[-2=-m-4;\qquad\qquad m=-2\]

\[R(t,y)=t^{-3}y^{-2}\]

Answer 13

applying the product rule was a mistake,

Answer 14

Oh that is okay..

I am still understanding it..

take me time to see how you solved it..

Answer 15

i havent solved it yet, but i have found the indicies of the integrating factor

Answer 16

Now what to do next??

Answer 17

\[f(t,y)=\int R(t,y)M\text dt\]

Answer 18

\[f(t,y)=\int t^{-2}y^{-3}(2t^3y-y^3)\text dt\]\[=\int \left(2ty^{-2}-t^{-2}\right)\text dt\]\[=\frac{t^2}{y^{2}}+\frac{1}{2t}+g(y)\]

Answer 19

\[f(t,y)=\int R(t,y)N\text dy\]

Answer 20

Let me try this..

Answer 21

Don't you think you are wrong somewhere??

Answer 22

if i am i cant see it ?
where are you looking?

Answer 23

\[\large R(t,y) = t^m y^n\]

Right ??

Answer 24

oh,

bother

Answer 25

\[\large R(t,y) = t^{-2} y^{-3}\]

Answer 26

got my alphabet backwords

Answer 27

thanks

Answer 28

Now you have to find f(t,y) again..

And I find f(t,y) for second one..

Answer 29

No no you have found them correct I guess..

Answer 30

have i got them wrong/?

Answer 31

\[f(t,y)=\int\limits R(t,y)N\text dy \implies f(t,y) = \int\limits(t^{-2}y^{-3})(-8t^3)dy\]

Answer 32

No..

In finding f(t,y) you have done right..

Answer 33

\[\implies -8t \int\limits (y^{-3})dy \implies f(t,y) = \frac{4t}{y^2} + g(t)\]

Answer 34

May be but I tried..

Answer 35

\[f(t,y)=\int R(t,y)M\text dt\]\[=\int t^{-2}y^{-3}(2t^3y-y^3)\text dt\]\[=\int \left(2ty^{-2}-t^{-2}\right)\text dt\]\[=\frac{-2t^2}{y^{2}}+\frac{1}{2t}+g(y)\]

\[f(t,y)=\int R(t,y)N\text dy\]\[=\int (t^{-2}y^{-3})(-2t^4) \text dy\]\[={-2}t^2\int y^{-3}\text dy\]\[=\frac{{-2}t^2}{y^{2}}+h(t)\]

\[g(y)=0;\qquad h(t)=\frac 1{2t}\]

\[f(t,y)=\frac{-2t^2}{y^{2}}+\frac{1}{2t}=c\]

Answer 36

Oh yeah I think integral has to taken first for finding N..

Answer 37

N is −2t^4

from the original equation

Answer 38

Yes..

Answer 39

we dont find N

Answer 40

I have found N by integral of dN/dt sorry..

Answer 41

\[=\int\limits (t^{-2}y^{-3})(-2t^4) \text dy \implies -2t^2 \int\limits y^{-3}dy = \frac{-2t^2}{(-2) \times y^2} = \frac{t^2}{y^2} + h(t)\]

Don't you think so ??

Answer 42

ok so i have made some mistakes

Answer 43

Mistakes are part of our lives..

Do it but try to do it once..

And after all we are now learners only..

Sometimes I cannot find mistakes of mine so Friends are always there to find your mistakes..

So, be friends forever so that you and I can find each other's mistake.

Ha ha ha ha..

Answer 44

its a deal

Answer 45

Okay as you said...

Answer 46

Ok my full solution;


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

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

\[\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial t}\]
\[R=R(t,y)=t^my^n\]
\[\frac{\partial MR(t,y)}{\partial y}=\frac{\partial NR(t,y)}{\partial t}\]\[\frac{\partial }{\partial y}\left((2t^3y-y^3)t^my^n\right)=\frac{\partial }{\partial t}\left((-2t^4)t^my^n\right)\]\[\frac{\partial }{\partial y}\left(2t^{m+3}y^{n+1}-t^my^{n+3}\right)=\frac{\partial }{\partial t}\left(-2t^{m+4}y^n\right)\]
\[2(n+1)t^{m+3}y^{n}-(n+3)t^my^{n+2}=-2(m+4)t^{m+3}y^n\]
\[2(n+1)=-2(m+4);\qquad\qquad (n+3)=0\]\[n+1=-m-4;\qquad\qquad n=-3\]\[-2=-m-4;\qquad\qquad m=-2\]
\[R=R(t,y)=\frac1{t^{2}y^{3}}\]
\[f(t,y)=\int R(t,y)M\text dt\]\[=\int \frac1{t^{2}y^{3}}(2t^3y-y^3)\text dt\]\[=\int \left(\frac{2t}{y^{2}}-\frac1{t^2}\right)\text dt\]\[=\frac{t^2}{y^{2}}+\frac{1}{t}+g(y)\]

\[f(t,y)=\int R(t,y)N\text dy\]\[=\int \frac1{t^{2}y^{3}}(-2t^4) \text dy\]\[={-2}t^2\int \frac 1{y^3}\text dy\]\[=\frac{t^2}{y^{2}}+h(t)\]
\[g(y)=0;\qquad\qquad h(t)=\frac 1t\]
\[f(t,y)=\frac{t^2}{y^{2}}+\frac{1}{t}=c\]

Answer 47

It is looking good to me..

How you put g(y) = 0 and h(t) = 1/t
any logic or procedure ??

Answer 48

i found \(g,h\) because \[f(t,y)=\int R(t,y)M\text dt=\int R(t,y)N\text dy\]\[\qquad\qquad\frac{t^2}{y^{2}}+\frac{1}{t}+g(y)=\frac{t^2}{y^{2}}+h(t)\]

comparing these

Answer 49

Oh I see..

Answer 50

the other method is to take the partial of derivative of say \[\frac{t^2}{y^{2}}+\frac{1}{t}+g(y)\] and to hav it equal N

Answer 51

I got this..
But now the solution remains or not ??

Answer 52

ah looks like the partial derivative should be equal to the integrating factor times N
\[\frac{\partial }{\partial y}\left(\frac{t^2}{y^{2}}+\frac{1}{t}+g(y)\right)=\frac{-2t^4}{t^{2}y^{3}}\]

\[-2\frac{t^2}{y^3}+g'(y)=\frac{-2t^2}{y^{3}}\]
\[g'(y)=0\]

Answer 53

\[g(y)=\int 0\cdot \text dy=c\]

Answer 54

yes..

Answer 55

\[\frac{\partial}{\partial t}\left(\frac{t^2}{y^{2}}+h(t)\right)=RM\]\[\frac{2t}{y^2}+h'(t)=\frac{2t^3y-y^3)}{t^{2}y^{3}}\]\[\frac{2t}{y^2}+h'(t)=\frac{2t}{y^2}-\frac 1{t^2}\]\[h'(t)=-\frac 1{t^2}\]\[h(t)=-\int \frac 1t\text dt\]\[h(t)=\frac 1t\]

Answer 56

\[\Large f(t,y)=\frac{t^2}{y^{2}}+\frac{1}{t}=c\]