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.
what does it mean by solve? find the function that it was derived from?
it might have something to do with it being an homogeneous function..... but I am just learning about them :) if its homogeneous, you can substitute y=vx into the equation and separate the variables.... when you get to the end of it, replace v by v=y/x. or something along those lines
Manipulate the equation so it is in Mdx+Ndy=0 form. To check if it's exact, partial M with respect to y has to equal partial N with respect to x. If they are equal, then you can apply the exact d.e method. Let F be another function where the solution to the d.e is F=c. Since the equation is exact, the following statements are true and will be applied. Partial F with respect to y equals M and partial F with respect to x equals N. The way I remember which N and M goes with which variable is to look at the dx and dy term. Since the first term is Mdx, it's partial F with respect to x since it corresponds to the dx. So now, since you have partial F with respect to x is equal to M. We can sub in our value of M since we know what it is. Our end goal is to find the function F, so we need to integrate with respect to x. Instead of just C, your constant will be a function of y, H(y) because you are integrating with respect to x. Now, to completely solve for the function F, you need H(y). So using your equation, partial F with respect to y equals N. Obtain partial F with respect to y by differentiating what you just got earlier with respect to y, keeping in mind your constant H(y) becomes H'(y). Now, equate this back to N. Rearrange using algebra to solve for H'(y). Integrate to get H(y). Sub H(y) back into your function F. Equate F=c and you are done.
yeah.... just like that :)
when you say partial M with respect to x, do you mean take the partial derivative of the m function with respect to x while holding the y's constant?
so if M(x,y)dx= 2xy-3x^2-2, dM/dx will be 2y-3?
Yeah that's right. An easier notation that is usually used is just Mx=2y-3x with the x being a subscript. Because dM/dx is different than the partial operator but I'm sure you know that.
i guess where I'm getting confused is when I test to see if it is exact which it should be, I am getting Mx=Ny, in this case, 2y-3x=12y which isn't a true statement
It's My=Mx, not other way around.
so I got two equations, df/dx=m=2xy-3x^2-2 and df/dy=N=-6y+x^2-3. I don't really follow what you were saying from here. do I integrate both of these, set them equal to eachother and solve for a y or an x?
No, you only need to integrate one. It doesn't matter which one you choose, but I always choose to integrate M. Our goal is to find F remember, so since df/dx=M you have to integrate with respect to x and then there's the whole thing about the constant y ou need to be careful of which I wrote above. Then you get the function F after integrating. You want to differentiate with respect to x this newly obtained F to get df/dx. You are doing this because you need to solve for the constant. Equate that to N, and solve for the constant you got earlier.
after integrating m, I got f(x,y)= ((2xy-2)*x)-x^3+c. if I differentiate this, won't I just get the same function that I integrated?
No because you are differentiating with respect to y now. Sorry I said x in the above post. And like I said, your constant is not just C. It is a function of y, H(y) because you integrated with respect to x, therefore any function of y also acts as a constant.