Question

Find f(x,y) such that (4x^3y^3-3x^2)dx+(3x^4y^2+cos(2y))dy=0

Answer 1

exact equations...

Answer 2

Are given above. I'm assuming I integrate... but I'm not sure of the proper technique

Answer 3

Usually you first rearrange the equation and then actually verify if it is exact or not. To be exact:
\[\Large M_y=N_x \]

Answer 4

So you are saying, rearrange the equation, integrate, then solve for x and y?

Answer 5

can we take antiderivative respect to x for the first part and the same respect to y for the second part and add them together to get f(x,y)?

Answer 6

or integral, the same process, right?

Answer 7

Sorry for my slow response, the website is acting super strange for me today, so I never really see what's going on. I said to rearrange it, this was a bit vague. The multivariable chain rules says:
\[\Large df(x,y)=f_xdy+f_ydy \]
So you can match that to your exact equation, is it exact?

\[\Large M_y=12x^3y^2 \\ \Large N_x=12x^3y^2 \\
\Large \therefore \ M_y=N_x \]
It is exact.

Answer 8

Meh, here again: Multivariable chainrule:
\[\Large d\left(f(x,y)\right)=f_xdx+f_ydy \]
that's much better.

Answer 9

Let me bring the stuff together:

\[\Large \Large d\left(f(x,y)\right) =f_xdx+f_ydy=0 \\
\Large(4x^ 3y^3-3x^2)dx+(3x^4y^2+\cos(2y))dy=0 \]

The first line introduces already that there is a solution:

\[\Large f(x,y)=C \]

For that to be true, we need to set:

\[\Large f_x=4x^3y^3-3x^2 \]

Integration leads to:

\[\Large f(x,y)=x^4y^3-x^3+g(y) \]

Notice that the g(y) is essential here, if you differentiate the above function, you will get f_x because g is a function of y and y only:

Differentiate the function to figure out what g(y) is, so partial differentiate with respect to y:
\[\Large f_y=3x^4y^2+g'(y)=3x^4y^2+\cos(2y) \]
Or \[\Large g(y)=\frac{1}{2}\sin(2y)=\sin y\cos y \]

Answer 10

Notice that I dropped another constant here after the integration for an explicit of g(y) because we already know by the primitive solution that it includes one constant. So we do not need another one. Combine the solutions and that's it:

\[\Large f(x,y)=x^4y^3-x^3+\sin (y)\cos (y)=C \]

Answer 11

hope that helps @bettirocked

Answer 12

@Spacelimbus why do we have to take several steps like that? why don't we just take integral respect to x, y respectively, it leads to the same answer, too. is there any reason else? I am not in that high, so I make some stupid question, please tell me.

Answer 13

There are no such things as stupid questions @Hoa, at least not when there is mathematics involved in it, it is tempting in such a problem to take the integrations respectively, and the answer is close:

\[\Large \int(4x^3y^3-3x^2)dx+\int(3x^4y^2+\cos(2y))dy=\int0  d(z)\]
So that would give us:

\[\Large x^4y^3-x^3+x^4y^3+\frac{1}{2}\sin(2y)=C  \]

which is equal to:

\[\Large 2x^4y^3-x^3+\sin(y)\cos(y)=C=f(x,y)\]

Not quite the same thing as above

Answer 14

The problem is mainly present in the standard definition of an exact equation. It is commonly written as:

\[\Large M(x,y)+N(x,y)\frac{dy}{dx}=0 \]
Which can also be arranged as:

\[\Large M(x,y)dx+N(x,y)dy=0 \]

The problem with this way of integration is the uniqueness which wont hold.

Answer 15

lol, exactly what happen in the situation of a 1-grade student tries to understand the material of a 5-grade student. too far from my mind, anyway, thanks a lot, I need time to get that high to understand what happen