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Revision difference : method of integrating factors |
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Version 3 |
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\PMlinkescapeword{equivalent} |
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| The {\em method of integrating factor} is in principle a means for solving ordinary differential equations of first \PMlinkescapetext{order}.\, It has not great practical significance, but is theoretically important. |
The {\em method of integrating factor} is in principle a means for solving ordinary differential equations of first \PMlinkescapetext{order}.\, It has not great practical significance, but is theoretically important. |
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Let's consider a differential equation solved for the derivative $y'$ of the unknown function and let's write the equation in the form
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Let's consider a differential equation solved in regard to the derivative and let's write in the form
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| \begin{align} |
\begin{align} |
| X(x,\,y)\,dx+Y(x,\,y)\,dy = 0. |
X(x,\,y)\,dx+Y(x,\,y)\,dy = 0. |
| \end{align} |
\end{align} |
| We assume that the functions $X$ and $Y$ have continuous partial derivatives in a region $R$ of $\mathbb{R}^2$. |
We assume that the functions $X$ and $Y$ have continuous partial derivatives in a region $R$ of $\mathbb{R}^2$. |
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| If there is a solution of (1) which may be expressed in the form |
If there is a solution of (1) which may be expressed in the form |
| $$f(x,\,y) = C$$ |
$$f(x,\,y) = C$$ |
| with $f$ having continuous partial derivatives in $R$ and with $C$ an arbitrary constant, then it's easy to see that such an $f$ satisfies the linear partial differential equation |
with $f$ having continuous partial derivatives in $R$ and with $C$ an arbitrary constant, then it's easy to see that such an $f$ satisfies the linear partial differential equation |
| \begin{align} |
\begin{align} |
| X\frac{\partial f}{\partial y}-Y\frac{\partial f}{\partial x} = 0. |
X\frac{\partial f}{\partial y}-Y\frac{\partial f}{\partial x} = 0. |
| \end{align} |
\end{align} |
| Conversely, every non-constant solution $f$ of (2) gives also a solution\, $f(x,\,y) = C$\, of (1).\, Thus, solving (1) and solving (2) are equivalent tasks. |
Conversely, every non-constant solution $f$ of (2) gives also a solution\, $f(x,\,y) = C$\, of (1).\, Thus, solving (1) and solving (2) are equivalent tasks. |
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| It's not difficult to show that if $f_0(x,\,y)$ is a non-constant solution of the equation (2), then all solutions of this equation are\, $F(f_0(x,\,y))$\, where F is a freely chosen function with (mostly) continuous derivative. |
It's not difficult to show that if $f_0(x,\,y)$ is a non-constant solution of the equation (2), then all solutions of this equation are\, $F(f_0(x,\,y))$\, where F is a freely chosen function with (mostly) continuous derivative. |
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| The connection of the equations (1) and (2) may be presented also in another form.\, Suppose that\, $f(x,\,y) = C$\, is any solution of (1).\, Then (2) implies the proportion equation |
The connection of the equations (1) and (2) may be presented also in another form.\, Suppose that\, $f(x,\,y) = C$\, is any solution of (1).\, Then (2) implies the proportion equation |
| $$\frac{f_x'}{X} = \frac{f_y'}{Y}.$$ |
$$\frac{f_x'}{X} = \frac{f_y'}{Y}.$$ |
| If we denote the common value of these two ratios by\, $\mu(x,\,y) = \mu$,\, then we have |
If we denote the common value of these two ratios by\, $\mu(x,\,y) = \mu$,\, then we have |
| $$f_x' = \mu X,\quad f_y' = \mu Y.$$ |
$$f_x' = \mu X,\quad f_y' = \mu Y.$$ |
| This gives to the differential of the function $f$ the expression |
This gives to the differential of the function $f$ the expression |
| $$d\,f(x,\,y) = \mu(x,\,y)(X(x,\,y)\,dx+Y(x,\,y)\,dy).$$ |
$$d\,f(x,\,y) = \mu(x,\,y)(X(x,\,y)\,dx+Y(x,\,y)\,dy).$$ |
| We see that\, $\mu(x,\,y)$\, is the {\em integrating factor} of the given differential equation (1), i.e. the left side of (1) turns, when multiplied by\, $\mu(x,\,y)$,\, to an exact differential. |
We see that\, $\mu(x,\,y)$\, is the {\em integrating factor} of the given differential equation (1), i.e. the left side of (1) turns, when multiplied by\, $\mu(x,\,y)$,\, to an exact differential. |
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| Conversely, any integrating factor $\mu$ of (1), i.e. such that\, |
Conversely, any integrating factor $\mu$ of (1), i.e. such that\, |
| $\mu Xdx+\mu Ydy)$\, is the differential of some function $f$, is easily seen to determine the solutions of the form\, $f(x,\,y) = C$\, of (1).\, Altogether, solving the differential equation (1) is equivalent with finding an integrating factor of the equation. |
$\mu\cdot(Xdx+Ydy)$\, is the differential of some function $f$, is easily seen to give the solutions of the form\, $f(x,\,y) = C$\, of (1).\, Altogether, solving the differential equation (1) is equivalent with finding an integrating factor of the equation. |
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| When an integrating factor $\mu$ of (1) is available, the solution function $f$ |
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| can be gotten from the line integral |
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| $$f(x,\,y) := \int_{P_0}^P [\mu(x,\,y)X(x,\,y)\,dx+\mu(x,\,y)Y(x,\,y)\,dy]$$ |
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| along any curve $\gamma$ connecting an arbitrarily chosen point\, $P_0 =(x_0,\,y_0)$\, and the point\, $P = (x,\,y)$\, in the region $R$. |
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