At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Hey 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, \[y+(2xy-e^{-2y})y'=0\]
\[ydx+(2xy-e^{-2y})dy=0\] \[M(x,y) = y \implies M_y = 1\]\[N(x,y)=(2xy-e^{-2y})dy \implies N_y = 2y\] 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, \[\frac{ M_y-N_x }{ M } = \frac{ 1-2y }{ y } \] 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 \[p(y)ydx+p(y)(2xy-e^{-2y})dy=0\] next I can look for my p(y) using the equation above, \[M(x,y)=p(y)y \implies M_y = p'(y)y+p(y)\]\[N(x,y) = p(y)(2xy-e^{2y}) \implies N_x = 2p(y)y\] where \[M_y=N_x\] then I get \[p'y=2py-p \implies p = \frac{ p'y }{ (2y-1) }\]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
Now I multiplied the equation but integrating factor\[\left( \frac{ e^{2y} }{ 2y } \right)ydx+\left( \frac{ e^{2y} }{ 2y } \right)(2xy-e^{-2y})dy=0\]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 \[g'(y) = -\frac{ 1 }{ 2y } \implies - \int\limits \frac{ 1 }{ 2y } dy \implies g(y) = -\frac{ 1 }{ 2 }\ln|y|+C\] which gives me \[\Psi(x,y) = \frac{ e^{2y} }{ 2 }x-\frac{ 1 }{ 2 }\ln|y|\] but the general solution is \[\Psi (x,y) = C \implies \frac{ e^{2y} }{ 2 }x-\frac{ 1 }{ 2 }\ln|y|=C\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

The solution does not have 1/2's, not sure where exactly my mistake is xD
That should say \[\psi_x = A ~~~\text{and}~~~\psi_y = B\]
So you can label your two functions anything? M and N or P and Q?
*Scratches head* I can't find the mistake..the answer should be \[xe^{2y}-\ln|y|=C\]
The M and N is just a way to find whether or not it's an exact equation
but yeah essentially, M(x,y)dx+N(x,y)dy = 0
Oh i see,kind of like test values, but instead of values you have functions.
I hate M and N crap, it's basically a waste of extra notation just cause 'it might not be exact' lol
Haha, I'm just doing it the way I learnt
Are you familiar with second derivatives commuting? \[\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]
I did that in calc 2, but omg you just gave me an idea to check my solution
\[f(x,y(x))=C\] 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: \[\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]
I 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.
multiply by 2 R.H.S=2C=C'
C'=C=A CONSTANT.
What do you mean @Empty Haha and if that's it...@surjithayer
\[xe ^{2y}-\ln y=2C=c\]
Thanks @surjithayer And yeah I see how that's done, but I don't know what you're trying to say @Empty
ok then
just 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
the M,N stuff is just how to see whether there exists an integrating factor in one-variable so we can reduce the problem to an ODE
$$y+(2xy-e^{-2y})y'=0\\y\, dx+(2xy-e^{-2y})dy=0$$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. $$d\Phi=\mu y\, dx+\mu\cdot(2xy-e^{-2y})dy$$so that it reduces to \(d\Phi=0\implies \Phi(x,y)=C\), giving the level sets of \(\Phi\) as our solution curves
Yes! 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.
if \(\mu y\, dx+\mu\cdot(2xy-e^{-2y})dy\) is the total differential of some \(\Phi\), then we'd expect $$\frac{\partial\Phi}{\partial x}=\mu y,\frac{\partial\Phi}{\partial y}=\mu\cdot(2xy-e^{-2y})$$and if these are partial derivatives of \(\Phi\) we expect symmetry of mixed partials, i.e. $$\frac{\partial^2\Phi}{\partial y\partial x}=\frac{\partial^2\Phi}{\partial x\partial y}\\\mu+\frac{\partial \mu}{\partial y}y=\mu\cdot2y+\frac{\partial\mu}{\partial x}\cdot (2xy-e^{-2y})$$by the product rule
now 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 \(\frac{d\mu}{dx}=\frac{\partial\mu}{\partial x}\) $$\mu=2\mu y+\frac{d\mu}{dx}(2xy-e^{-2y})\\\frac{d\mu}{dx}+\frac{2y-1}{2xy-e^{-2y}}\mu=0$$ 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: $$\mu+\frac{d\mu}{dy}=2y\mu$$
oops, 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 $$y\frac{d\mu}{dy}+(1-2y)\mu=0\\\frac1\mu\frac{d\mu}{dy}=\frac{2y-1}y\\\int\frac1\mu\frac{d\mu}{dy}dy=\int\left(2-\frac1y\right)\,dy\\\ln\mu=2y-\ln y+C\\\mu=Ce^{2y}/y$$
so this tells us that if we multiply throughout the equation by \(\mu\), we get a total differential of some function \(\Phi\), so $$\frac{d\Phi}{dx}=\mu y=Ce^{2y}\\\frac{d\Phi}{dy}=\mu\cdot(2xy-e^{-2y})=2Cxe^{2y}-C/y$$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
now notice if we move them all to one side: $$xe^{2y}-\ln y=\frac{C_3-C_2}{C_1}=C$$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: $$\Phi=A\implies \Phi+C=A+C$$
and 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).
You answered all my questions even ones I hadn't asked, thanks a lot @oldrin.bataku that was great!

Not the answer you are looking for?

Search for more explanations.

Ask your own question