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Astrophysics
 one year ago
@ganeshie8
Astrophysics
 one year ago
@ganeshie8

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Hey ganeshie could you just check over my work, I want to know if I'm doing this right, i have to find an integrating factor and solve the given equation, it'll take me a bit to type all this out haha, \[y+(2xye^{2y})y'=0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3\[ydx+(2xye^{2y})dy=0\] \[M(x,y) = y \implies M_y = 1\]\[N(x,y)=(2xye^{2y})dy \implies N_y = 2y\] so this is obviously not an exact equation as\[M_y \neq Nx\] So I used the following to see whether to find a integrating factor that depends on y or x, \[\frac{ M_yN_x }{ M } = \frac{ 12y }{ y } \] so this depends on only y, that means I can use this p = p(y). Now I multiply the entire equation by p(y) giving me \[p(y)ydx+p(y)(2xye^{2y})dy=0\] next I can look for my p(y) using the equation above, \[M(x,y)=p(y)y \implies M_y = p'(y)y+p(y)\]\[N(x,y) = p(y)(2xye^{2y}) \implies N_x = 2p(y)y\] where \[M_y=N_x\] then I get \[p'y=2pyp \implies p = \frac{ p'y }{ (2y1) }\]doing integrating I got \[p= e^{2y}e^{1}e^c \frac{ 1 }{ 2y }\] and then I replaced \[\frac{ e^c }{ e } = c_1\] I'm not exactly sure why we don't include the constants but then my integrating factor is \[p(y) = \frac{ e^{2y} }{ 2y }\] ...haha this is pretty long

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Now I multiplied the equation but integrating factor\[\left( \frac{ e^{2y} }{ 2y } \right)ydx+\left( \frac{ e^{2y} }{ 2y } \right)(2xye^{2y})dy=0\]which simplifies to\[\frac{ e^{2y} }{ 2 }dx+(e^{2y}x\frac{ 1 }{ 2y })dy=0\] now we check to see if it's an exact equation I got\[M(x,y)=\frac{ e^{2y} }{ 2 } \implies M_y = e^{2y}\] and\[N(x,y) = e^{2y}x\frac{ 1 }{ 2y } \implies N_x = e^{2y}\] therefore \[M_y=N_x \implies \text{exact}\] \[\psi_x = M(x,y) = \frac{ e^{2y} }{ 2 }\]\[\psi_y = N(x,y) = e^{2y}x\frac{ 1 }{ 2y }\] letting \[\psi_1 = A~~~\text{and}~~~\psi_2=B\] now I'll integrate A \[\psi_x = \frac{ e^{2y} }{ 2 } \implies \int\limits \frac{ e^{2y} }{ 2 }dx = \frac{ e^{2y} }{ 2 }x+g(y) = \Psi(x,y)\] now I took the derivative with respect to y, \[\frac{ \partial (\frac{ e^{2y} }{ 2 }x+g(y)) }{ \partial y } = e^{2y}x+g'(y)\] comparing with B \[g'(y) = \frac{ 1 }{ 2y } \implies  \int\limits \frac{ 1 }{ 2y } dy \implies g(y) = \frac{ 1 }{ 2 }\lny+C\] which gives me \[\Psi(x,y) = \frac{ e^{2y} }{ 2 }x\frac{ 1 }{ 2 }\lny\] but the general solution is \[\Psi (x,y) = C \implies \frac{ e^{2y} }{ 2 }x\frac{ 1 }{ 2 }\lny=C\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3The solution does not have 1/2's, not sure where exactly my mistake is xD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3That should say \[\psi_x = A ~~~\text{and}~~~\psi_y = B\]

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0So you can label your two functions anything? M and N or P and Q?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3*Scratches head* I can't find the mistake..the answer should be \[xe^{2y}\lny=C\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3The M and N is just a way to find whether or not it's an exact equation

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3but yeah essentially, M(x,y)dx+N(x,y)dy = 0

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.0Oh i see,kind of like test values, but instead of values you have functions.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I hate M and N crap, it's basically a waste of extra notation just cause 'it might not be exact' lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Haha, I'm just doing it the way I learnt

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Are you familiar with second derivatives commuting? \[\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3I did that in calc 2, but omg you just gave me an idea to check my solution

Empty
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x,y(x))=C\] Chain rule: \[\frac{d f}{dx} = \frac{\partial f}{\partial x} \frac{dx}{dx} + \frac{\partial f}{\partial y} \frac{dy}{dx} = 0\] Ok really \(\frac{dx}{dx}=1\) so throw this away to get this: \[\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}y' = 0\] Now we see that if something of this form is truly exact, then clearly: \[\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I mean you sorta explained this earlier now that I'm scrolling up so I'm kinda frightened by the fact that you said you didn't know what that meant that the second derivatives commute like that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiply by 2 R.H.S=2C=C'

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3What do you mean @Empty Haha and if that's it...@surjithayer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[xe ^{2y}\ln y=2C=c\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Thanks @surjithayer And yeah I see how that's done, but I don't know what you're trying to say @Empty

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just multiply through by \(2\)  the whole point is that solutions to an exact differential equation are level sets of some potential function \(\Phi\), i.e. $$\Phi=C$$ whether we multiply both sides by \(2)\) doesn't make a difference, since $$2\Phi=2C\Leftrightarrow\Phi=C$$ gives the same curve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the M,N stuff is just how to see whether there exists an integrating factor in onevariable so we can reduce the problem to an ODE

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$y+(2xye^{2y})y'=0\\y\, dx+(2xye^{2y})dy=0$$so we want to know if there exists some integrating factor \(\mu\) such that we can write the LHS as the total differential of some potnetial function \(\Phi\), i.e. $$d\Phi=\mu y\, dx+\mu\cdot(2xye^{2y})dy$$so that it reduces to \(d\Phi=0\implies \Phi(x,y)=C\), giving the level sets of \(\Phi\) as our solution curves

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3Yes! Thanks @oldrin.bataku exactly what I was looking for, I realized when I first started exact equations they give a solution to a potential function, or are one in the same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if \(\mu y\, dx+\mu\cdot(2xye^{2y})dy\) is the total differential of some \(\Phi\), then we'd expect $$\frac{\partial\Phi}{\partial x}=\mu y,\frac{\partial\Phi}{\partial y}=\mu\cdot(2xye^{2y})$$and if these are partial derivatives of \(\Phi\) we expect symmetry of mixed partials, i.e. $$\frac{\partial^2\Phi}{\partial y\partial x}=\frac{\partial^2\Phi}{\partial x\partial y}\\\mu+\frac{\partial \mu}{\partial y}y=\mu\cdot2y+\frac{\partial\mu}{\partial x}\cdot (2xye^{2y})$$by the product rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we're interested in seeing whether this reduces to a nice ODE if we presume, say: \(\mu=\mu(x)\), i.e. \(\frac{\partial\mu}{\partial y}=0\) and \(\frac{d\mu}{dx}=\frac{\partial\mu}{\partial x}\) $$\mu=2\mu y+\frac{d\mu}{dx}(2xye^{2y})\\\frac{d\mu}{dx}+\frac{2y1}{2xye^{2y}}\mu=0$$ or if we consider \(\mu=\mu(y)\) so \(\frac{\partial\mu}{\partial x}=0,\frac{\partial\mu}{\partial\mu}=\frac{d\mu}{dy}\) and try the other way: $$\mu+\frac{d\mu}{dy}=2y\mu$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, i meant to write $$\mu+y\frac{d\mu}{dy}=2y\mu$$ anyways, you don't have to be a genius to see one of those is much easier to solve than the other :) so consider $$y\frac{d\mu}{dy}+(12y)\mu=0\\\frac1\mu\frac{d\mu}{dy}=\frac{2y1}y\\\int\frac1\mu\frac{d\mu}{dy}dy=\int\left(2\frac1y\right)\,dy\\\ln\mu=2y\ln y+C\\\mu=Ce^{2y}/y$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this tells us that if we multiply throughout the equation by \(\mu\), we get a total differential of some function \(\Phi\), so $$\frac{d\Phi}{dx}=\mu y=Ce^{2y}\\\frac{d\Phi}{dy}=\mu\cdot(2xye^{2y})=2Cxe^{2y}C/y$$integrating gives us $$\Phi=\int Ce^{2y}\, dx=Cxe^{2y}+g(y)\\\Phi=\int (2Cxe^{2y}C/y)\,dy=Cxe^{2y}C\ln y+f(x)\\\implies\Phi(x,y)=C_1(xe^{2y}\ln y)+C_3$$ so our solutions are therefore level sets of \(\Phi\), i.e. $$C(xe^{2y}\ln y)+C_2=C_3$$ notice we have what seems to be three arbitrary constants

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now notice if we move them all to one side: $$xe^{2y}\ln y=\frac{C_3C_2}{C_1}=C$$and those arbitrary constants combine to give (drum roll...) another arbitrary constant :) notice that we're free to pick any scalar multiple of our integration factor simply because if \(M\, dx+N\, dy=0\) has potential function \(\Phi\), then \(CM\, dx+CN\, dy=0\) has potential function \(C\cdot\Phi\) and the two solutions are 'really' the same: $$\Phi=A\implies C\cdot\Phi=CA$$ similarly, when integrating there are actually infinitely many valid potential functions up to a constant simply because the constant terms are lost when we take our total differential and once again the solutions are 'really' the same: $$\Phi=A\implies \Phi+C=A+C$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then one of the arbitrary constants is actually a parameter for the family of solution curves to our problem, because *every* level set of \(\Phi\) satisfies \(d\Phi=0\). to pick a particular solution, we need some more information (like an initial value), so there's really only one degree of freedom here (the other two constants of integration we found just make our family of solutions look superficially different).

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.3You answered all my questions even ones I hadn't asked, thanks a lot @oldrin.bataku that was great!
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