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Determine whether equation is exact.

alane1994

Active member
Oct 16, 2012
126
I have several examples in my book for this problem, but I don't understand what they are saying. I guess I don't get the format. I will post up the example I think applies to the problem, and if one of you could explain the example that would be fantastic.

Determine whether each of the equations is exact. If it is exact, find the solution.
\(3x^2-2xy+2)+(6y^2-x^2+3)y\prime=0\)
This is my problem.

Example Problem.
Solve the differential equation
\[2x+y^2+2xyy\prime=0\]
The equation is neither linear nor separable, so the methods suitable for those types of equations are not applicable here. However observe that the function \(\psi(x,y)=x^2+xy^2\) has the property that
\[2x+y^2=\dfrac{\partial\psi}{\partial x},~~~2xy=\dfrac{\partial\psi}{\partial y}\]
Therefore, the differential equation can be written as
\[\dfrac{\partial\psi}{\partial x} +\dfrac{\partial\psi}{\partial y}\dfrac{dy}{dx}=0\]
Assuming that y is a function of x, we can use the chain rule to write the left side of Eq.3 as \(\dfrac{d\psi(x,y)}{dx}\). Then Eq.3 has the form
\[\dfrac{d\psi}{dx}(x,y)=\dfrac{d}{dx}(x^2+xy^2)=0\]
By integrating Eq.4 we obtain
\[\psi(x,y)=x^2+xy^2=c\]
where c is an arbitrary constant. the level curves of \(\psi(x,y)\) are the integral of curves of Eq.1. Solutions of Eq.1 are defined implicitly by Eq.5.



My main questions are what is with the notation of \(\text{arbitrary holder}(x,y)\)?
And, just some clarification would be fantastic.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,893
Hi alane!

Zooming in on your example problem first.

Note that when you integrate $2x+y^2$ with respect to x, you get $x^2 +xy^2 + C(y)$.
I have written $C(y)$ here to indicate a function that only depends on y.
It represents the usual integration constant that vanishes when differentiating with respect to x.

And when we integrate $2xy$ with respect to y, we get $xy^2 + D(x)$, where $D(x)$ is the integration constant that can still be a function of x, but not of y.

See the similarity?
Can you perhaps find functions $C(y)$ and $D(x)$ such that these 2 expressions become the same?
If so, let's call it $\psi(x,y)$ and look back to how it relates to the original differential equation...
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
I have several examples in my book for this problem, but I don't understand what they are saying. I guess I don't get the format. I will post up the example I think applies to the problem, and if one of you could explain the example that would be fantastic.

Determine whether each of the equations is exact. If it is exact, find the solution.

\(3x^2-2xy+2)+(6y^2-x^2+3)y\prime=0\)

This is my problem.
A first order ODE in the form...

$\displaystyle y^{ '}= - \frac{A(x,y)}{B(x,y)}\ (1)$

... is called 'exact' if ...

$\displaystyle \frac{\partial A}{\partial y}= \frac{\partial B}{\partial x}\ (2)$

In this case the solution of (1) is...

$\displaystyle \int_{x_{0}}^{x} A (u,y)\ du + \int_{y_{0}}^{y} B(x,v)\ dv = c\ (3)$

... where $(x_{0}, y_{0}$ is a point where A and B are defined and c is an arbitrary constant. In Your case is $\displaystyle A = 3 x^{2} - 2 x y + 2$, $B = 6 y^{2} - x^{2} + 3$ and the (2) is verified, so that the (3) becomes...

$\displaystyle \int_{0}^{x} (3 u^{2} - 2 u y +2)\ du + \int_{0}^{y} (6 v^{2} - x^{2} + 3)\ dv = c\ (4)$

The integration of (4) is left to You...

Kind regards

$\chi$ $\sigma$