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<math> | <math> | ||
| − | M(x,y)+N(x,y)y' = 0 | + | M(x,y)+N(x,y)y' = 0\qquad \qquad (*) |
</math> | </math> | ||
| Line 29: | Line 29: | ||
\psi_x(x,y)=M(x,y);\qquad \psi_y(x,y)=N(x,y) | \psi_x(x,y)=M(x,y);\qquad \psi_y(x,y)=N(x,y) | ||
</math> | </math> | ||
| + | |||
| + | |||
| + | == Example == | ||
| + | |||
| + | Suppose your differential equation is | ||
| + | |||
| + | <math> | ||
| + | y\,cosx+2xe^y+y'sinx+y'x^2e^y=y' | ||
| + | </math> | ||
| + | |||
| + | First, rearrange the terms so that they are in the same format as (*). | ||
| + | |||
| + | <math> | ||
| + | (y\,cosx+2xe^y)+(sinx+x^2e^y-1)y'=0 | ||
| + | </math> | ||
| + | |||
| + | First, make sure that M<sub>y</sub> = N<sub>x</sub>. Differentiating, we obtain: | ||
| + | |||
| + | <math> | ||
| + | M_y=cosx+2xe^y=N_x | ||
| + | </math> | ||
| + | |||
| + | Therefore ψ(x,y) exists. To find it, we use the fact that ψ<sub>x</sub>(x,y) = M(x,y). This implies that we should integrate M(x,y) with respect to x. Let's do so now. | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | \psi_x(x,y)&=M(x,y)\\ | ||
| + | \psi_x(x,y) &= y\,cosx+2xe^y\\ | ||
| + | \int \psi_x(x,y)\,dx &= \int y\,cosx+2xe^ydx\\ | ||
| + | \psi(x,y)&=y\,sinx+x^2e^y+h(y) | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Rather than obtaining an ordinary constant of integration, we obtain an arbitrary function of y, which we call h(y). The reason for this is that ψ(x,y) is a function of both x and y. Normally, we add the arbitrary constant C because we know that differentiation gets rid of it. Here, any function of just y is similarly going to be taken out by differentiation with respect to x. | ||
| + | |||
| + | The next step is to use the fact that ψ<sub>y</sub>(x,y)=N(x,y). Therefore, differentiate the last line of our previous calculation with respect to y, and set this equal to N(x,y). | ||
| + | |||
| + | <math> | ||
| + | \frac{\partial}{\partial y}\psi(x,y)=\frac{\partial}{\partial y}(ysinx+x^2e^y+h(y))=\frac{\partial}{\partial y}(sinx+x^2e^y-1) | ||
| + | </math> | ||
| + | <math> | ||
| + | \psi_y(x,y)=sinx+x^2e^y+h'(y)=x^2e^y | ||
| + | </math> | ||
| + | |||
| + | (this page is under construction) | ||
<nowiki>*</nowiki> There is another condition required for this technique to work that I did not want to mention initially for continuity reasons. The functions M, N, ∂M/∂y and ∂N/∂x must be continuous in a simply connected region R of the xy plane, and it must be true for every point in R that ∂M/∂y=∂N/∂x. By "simply connected", I simply mean that there are no holes in the region. Now, you should know this requirement, but most likely your professor will not try to trick you into using the technique mentioned on this page by giving you a problem in which R is not simply connected. So just take notice of this requirement and shelve it somewhere in your mind. | <nowiki>*</nowiki> There is another condition required for this technique to work that I did not want to mention initially for continuity reasons. The functions M, N, ∂M/∂y and ∂N/∂x must be continuous in a simply connected region R of the xy plane, and it must be true for every point in R that ∂M/∂y=∂N/∂x. By "simply connected", I simply mean that there are no holes in the region. Now, you should know this requirement, but most likely your professor will not try to trick you into using the technique mentioned on this page by giving you a problem in which R is not simply connected. So just take notice of this requirement and shelve it somewhere in your mind. | ||
Revision as of 15:09, 11 October 2009
Exact Equations
Before we begin, a quick note on notation. Within this section, subscripts of x and y mean "partial derivative with respect to x" and "partial derivative with respect to y" respectively. So
$ M_y(x,y) $
means the partial derivative of M(x,y) with respect to y.
Suppose, firstly, that your differential equation can be written this way:
$ M(x,y)+N(x,y)y' = 0\qquad \qquad (*) $
and secondly, that
$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $
In this case*, your differential equation is said to be exact, and the solution to the differential equation is ψ(x,y) where
$ \psi_x(x,y)=M(x,y);\qquad \psi_y(x,y)=N(x,y) $
Example
Suppose your differential equation is
$ y\,cosx+2xe^y+y'sinx+y'x^2e^y=y' $
First, rearrange the terms so that they are in the same format as (*).
$ (y\,cosx+2xe^y)+(sinx+x^2e^y-1)y'=0 $
First, make sure that My = Nx. Differentiating, we obtain:
$ M_y=cosx+2xe^y=N_x $
Therefore ψ(x,y) exists. To find it, we use the fact that ψx(x,y) = M(x,y). This implies that we should integrate M(x,y) with respect to x. Let's do so now.
$ \begin{align} \psi_x(x,y)&=M(x,y)\\ \psi_x(x,y) &= y\,cosx+2xe^y\\ \int \psi_x(x,y)\,dx &= \int y\,cosx+2xe^ydx\\ \psi(x,y)&=y\,sinx+x^2e^y+h(y) \end{align} $
Rather than obtaining an ordinary constant of integration, we obtain an arbitrary function of y, which we call h(y). The reason for this is that ψ(x,y) is a function of both x and y. Normally, we add the arbitrary constant C because we know that differentiation gets rid of it. Here, any function of just y is similarly going to be taken out by differentiation with respect to x.
The next step is to use the fact that ψy(x,y)=N(x,y). Therefore, differentiate the last line of our previous calculation with respect to y, and set this equal to N(x,y).
$ \frac{\partial}{\partial y}\psi(x,y)=\frac{\partial}{\partial y}(ysinx+x^2e^y+h(y))=\frac{\partial}{\partial y}(sinx+x^2e^y-1) $ $ \psi_y(x,y)=sinx+x^2e^y+h'(y)=x^2e^y $
(this page is under construction)
* There is another condition required for this technique to work that I did not want to mention initially for continuity reasons. The functions M, N, ∂M/∂y and ∂N/∂x must be continuous in a simply connected region R of the xy plane, and it must be true for every point in R that ∂M/∂y=∂N/∂x. By "simply connected", I simply mean that there are no holes in the region. Now, you should know this requirement, but most likely your professor will not try to trick you into using the technique mentioned on this page by giving you a problem in which R is not simply connected. So just take notice of this requirement and shelve it somewhere in your mind.
