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I've only begun to study exact equations so this question will be naive.

I've went through the process of taking a given differential equation, showing that it's an exact equation and hence the derivative of some function $\psi$. Figuring out the function $\psi$ is supposed to be a solution but I don't understand what is actually solved. If I know $\psi$, what information do I have that I didn't have when I only had the differential equation? What can I do with $\psi$?

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    You can evaluate it at a point. This lets you predict the behavior e.g. of a physical system at some time $t$.2012-10-22

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Let's say that you have the equation $ f(x,y)dx+g(x,y)dy=0 $ and that it is exact. Then you find $\psi(x,y)$ such that $ \frac{\partial \psi}{\partial x}=f,\quad\frac{\partial \psi}{\partial y}=g. $ The solution is given in implicit form as $ \psi(x,y)=C. $ What does this mean? That from the implicit solution you can solve for $y$ in terms of $x$, and that the correspondig function $y=y(x,C)$ is a solution of the differential equation.

Example: $ 2xdx+2ydy=0 $ is exact. The solution is given in implicit form as $ x^2+y^2=C. $ From it you can get explicit solutions: $ y(x)=\sqrt{C-x^2},\quad C>0,\quad |x|<\sqrt C. $ This last step is not always possible (meaning that although a solution $y=y(x,C)$ exists, you may not be able to find an explicit solution in terms of elementary functions.)

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Hint: Let you have an equation like $y'=f(x,y)$ which can be written as $y'=-\frac{M(x,y)}{N(x,y)}$ for some two functions $M(x,y)$ and $N(x,y)$. So you have $\frac{dy}{dx}=-\frac{M(x,y)}{N(x,y)}$ or $M(x,y)dx+N(x,y)dy=0$ If you can write the LHS of the equation as a differential of a certain $\phi$ such that $d\phi=M(x,y)dx+N(x,y)dy$, then by integration of both sides you get $d\phi=0\Longrightarrow\phi=C$ for some constant $C$.