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#### alane1994

##### Active member

- Oct 16, 2012

- 126

**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.