Solving a Homework Equation: (3xy^2+4y)dx+(3x^2y+4x)dy=0

[mshiddensecret]

Homework Statement

(3xy^2+4y)dx+(3x^2y+4x)dy=0

Homework Equations

The Attempt at a Solution

So First I checked if both equations were exact. I took the derivative of 3xy^2+4y and also derivative of the other and they were both equal so the equation is exact.

I took the 3xy^2+4y and integrated it with respect to x. And then I differentiated it to y.

I got 3x^2y+4x out of it. The problem is that it is exactly the same as the one from the originally equation so I did not ending up finding C.

Did I do something wrong?

 

[Mark44]

Homework Statement

(3xy^2+4y)dx+(3x^2y+4x)dy=0

Homework Equations

The Attempt at a Solution

So First I checked if both equations were exact. I took the derivative of 3xy^2+4y and also derivative of the other and they were both equal so the equation is exact.

I took the 3xy^2+4y and integrated it with respect to x. And then I differentiated it to y.

I got 3x^2y+4x out of it. The problem is that it is exactly the same as the one from the originally equation so I did not ending up finding C.

Did I do something wrong?

I agree that the equation is exact. Where you went wrong is in your integration.
$$\int (3xy^2 + 4y)dx \neq 3x^2y + 4x$$
When you integrate the above, treat y as if it were a constant.
Do a similar integration, with respect to y, for the other part.

Your textbook should have some examples of how this works.

 

[mshiddensecret]

No after I integrated, I got 3/2x^2y^2+4xy

Then differentiate and got 3x^2y+4x

 

[Brian T]

The function you are looking for is the one after you integrated. If you take the derivative and you get back to that function, you do not need to account for any extra functions. It looks correct.

∫(F_x)dx = F(x,y) + F(y) + C ... (1)
∫(F_y)dy = F(x,y) + F(x) + C ... (2)

In this case, if you integrate (1), take the derivative with respect to y, and compare it to F_y that's given, you should be able to determine any term that was missing. In this case, there are no terms to account for

 

[mshiddensecret]

So C = 0?

 

[ehild]

No after I integrated, I got

Then differentiate and got 3x^2y+4x

You miss the integration constant, which can be function of y. So the integral is 3/2x^2y^2+4xy+C(y).
Differentiate it with respect y. It is 3x^2y+4x+C'. It must be equal to 3x^2y+4x, so C' = 0. It means that C is a ?

 

[Brian T]

∫(Fx)dx = Φ(x,y) + F(y) + C ... (1)
∫(Fy)dy = Φ(x,y) + F(x) + C ... (2)
F(y) and F(x) are zero. C is just C, the constant of integration (a number and not a function of x or y). There is a distinction between functions of y or x that arise from integration w.r.t the other variable, and the standard integration constant that arises from any integration. The former is zero, while the latter is to be left undetermined.

The solution to an exact equation is written in the form:
Φ(x,y) + f(x) + f(y) = C. In this case, f(x) and f(y) are just zero, so:
Φ(x,y) = C should be the answer

 

[epenguin]

Has a speck of dust surrounded my brain like sometimes, or are you meant to notice that the original factorises such that it reduces to a distinctly easier problem than it is looking?

 

[Chestermiller]

$d(x^2y^2)=?$
$d(xy)=?$

Chet

 

[epenguin]

Too clever and unnecessary IMHO.

 

[epenguin]

No one? Won't the OP look at the two expressions, see that they obviously have a factor, that leads to solving the problem in one or two lines?

 

[epenguin]

Homework Statement

(3xy^2+4y)dx+(3x^2y+4x)dy=0

Well my hint has been up for several days now.

Isn't this just

(3xy + 4)(y dx + x dy) = 0

so that all you have to solve is

dy/y = - dx/x

?

 

[chwala]

No after I integrated, I got 3/2x^2y^2+4xy

Then differentiate and got 3x^2y+4x

After integration, you missed to include the constant which is a function of $y$, then follow the steps from there...that is the missing link.

 

[chwala]

Well my hint has been up for several days now.

Isn't this just

(3xy + 4)(y dx + x dy) = 0

so that all you have to solve is

dy/y = - dx/x

?

there is a factor that will cancel out...to realize that...wow, this post is a bit old...2015

 

[chwala]

You miss the integration constant, which can be function of y. So the integral is 3/2x^2y^2+4xy+C(y).
Differentiate it with respect y. It is 3x^2y+4x+C'. It must be equal to 3x^2y+4x, so C' = 0. It means that C is a ?

are the two solutions both acceptable? ie post $6$ and $12$?

 

[fresh_42]

are the two solutions both acceptable? ie post $6$ and $12$?

The main problem is, that we are not told whether $y$ is a function of $x,$ or both describe a curve in $\mathbb{R}^2.$ Post #6 is correct, #12 has the problem that it neglects the solution $3xy+4=0.$ And neither copes with the case that $y=y(t),x=x(t)$ could be a parameterized curve, in which case we would need an integration path.

If $y=y(x)$ then the original equation should be written as
\begin{align*}
0&=(4 y(x) + 3 x y(x)^2) dx + (4 x + 3 x^2 y(x)) dy = (3xy(x) + 4)(y(x) dx + x dy)
\end{align*}

However, if $3xy(x)+4=0$, then we get $y(x) dx + x dy=0$ automatically. Post #6 should yield the same result.

 

[chwala]

Is this particular differential equation homogenous?
Considering $M=3xy^2+4y$
$N= 3x^2y+4x$

and using,
$f(tx,ty)=t^nf(x,y)$...equation $1$

My steps are as follows:
On considering $M$, i get;
$f(x,y)=3xy^2+4y$
$f(tx,ty)=3tx(ty)^2+4ty$
=$3t^3xy^2+4ty$
and considering $N$, i get;
$f(x,y)=3x^2y+4x$
$f(tx,ty)=3(tx)^2ty+4tx$
= $3t^3x^2y+4tx$
now from the steps, i cannot see how we can factor out $t^n$...

oooooh, unless we are going to have,

$\frac {ty(3t^2xy+4)} {tx (3t^2xy+4)}$

=$\frac {t(t^2(3xy^2)+4y)} {t (t^2(3x^2y)+4)}$...but it does not satisfy equation $1$...

 

[fresh_42]

It isn't homogeneous because $\deg 3x^2y =\deg 3xy^2 = 3$ and $\deg 4x =\deg 4y =1.$ Plus, I don't think that $x,y$ are meant to be both variables, although the original equation does look as if they were, because we don't have a parameterized integration path. Where should we integrate if it was a level surface in $\mathbb{R}^3\,\rm ?$

The solutions are most likely simply $y=-\dfrac{3}{4x}$ or $y=\dfrac{c}{x}.$

 

[chwala]

It isn't homogeneous because $\deg 3x^2y =\deg 3xy^2 = 3$ and $\deg 4x =\deg 4y =1.$ Plus, I don't think that $x,y$ are meant to be both variables, although the original equation does look as if they were, because we don't have a parameterized integration path. Where should we integrate if it was a level surface in $\mathbb{R}^3\,\rm ?$

The solutions are most likely simply $y=-\dfrac{3}{4x}$ or $y=\dfrac{c}{x}.$

Noted fresh, bingo!

 

[chwala]

so can we have non homogenous (first order differential equations) being solved by separation of variables? ie by using the substitution, $y=vx$?

 

[fresh_42]

so can we have non homogenous (first order differential equations) being solved by separation of variables? ie by using the substitution, $y=vx$?

Maybe, but here we have the solution $y=vx^{-1}$ so $y=vx$ won't help.

If you want to see an example where $x$ and $y$ are variables, and the differential form is integrated along a certain path, have a look at:
https://www.physicsforums.com/threads/why-the-terms-exterior-closed-exact.871875/#post-5474443

 

[chwala]

Maybe, but here we have the solution $y=vx^{-1}$ so $y=vx$ won't help.

If you want to see an example where $x$ and $y$ are variables, and the differential form is integrated along a certain path, have a look at:
https://www.physicsforums.com/threads/why-the-terms-exterior-closed-exact.871875/#post-5474443

I will check this out...