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Abel's Identity

Given a homogeneous linear Second-Order Ordinary Differential Equation,

\end{displaymath} (1)

call the two linearly independent solutions $y_1(x)$ and $y_2(x)$. Then
\end{displaymath} (2)

\end{displaymath} (3)

Now, take $y_1 \times$ (3) minus $y_2 \times$ (2),

\end{displaymath} (4)

\end{displaymath} (5)

\end{displaymath} (6)

Now, use the definition of the Wronskian and take its Derivative,
$\displaystyle W$ $\textstyle \equiv$ $\displaystyle y_1y'_2-y'_1y_2$ (7)
$\displaystyle W'$ $\textstyle =$ $\displaystyle (y'_1y'_2+y_1y''_2)-(y'_1y'_2+y''_1y_2)$  
  $\textstyle =$ $\displaystyle y_1y''_2-y''_1y_2.$ (8)

Plugging $W$ and $W'$ into (6) gives
W'+PW = 0.
\end{displaymath} (9)

This can be rearranged to yield
{dW\over W} = -P(x)\,dx
\end{displaymath} (10)

which can then be directly integrated to
\ln\left[{W(x)\over W_0}\right]= - \int P(x)\,dx,
\end{displaymath} (11)

where $\ln x$ is the Natural Logarithm. Exponentiation then yields Abel's identity
W(x) = W_0e^{-\int P(x)\,dx},
\end{displaymath} (12)

where $W_0$ is a constant of integration.

See also Ordinary Differential Equation--Second-Order


Boyce, W. E. and DiPrima, R. C. Elementary Differential Equations and Boundary Value Problems, 4th ed. New York: Wiley, pp. 118, 262, 277, and 355, 1986.

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© 1996-9 Eric W. Weisstein