- #1
DivGradCurl
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An exact equation has the form
[tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]
where
[tex]M(x,y) = \frac{\partial \psi}{\partial x} (x,y)[/tex]
and
[tex]N(x,y) = \frac{\partial \psi}{\partial y} (x,y) \mbox{.}[/tex]
If [tex]y=\phi (x)[/tex] and [tex]\psi (x,y) = c[/tex], then
[tex]M(x,y) + N(x,y) \: y^{\prime} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right] = 0 \mbox{.}[/tex]
I can't understand this:
[tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]
Any help is highly appreciated.
[tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]
where
[tex]M(x,y) = \frac{\partial \psi}{\partial x} (x,y)[/tex]
and
[tex]N(x,y) = \frac{\partial \psi}{\partial y} (x,y) \mbox{.}[/tex]
If [tex]y=\phi (x)[/tex] and [tex]\psi (x,y) = c[/tex], then
[tex]M(x,y) + N(x,y) \: y^{\prime} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right] = 0 \mbox{.}[/tex]
I can't understand this:
[tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]
Any help is highly appreciated.