Abel's Identity

$\begin{displaymath} y''+P(x)y'+Q(x)y=0, \end{displaymath}$

(1)

call the two linearly independent solutions

and

. Then

$\begin{displaymath} y''_1+P(x)y'_1+Q(x)y_1=0 \end{displaymath}$

(2)

$\begin{displaymath} y''_2+P(x)y'_2+Q(x)y_2=0. \end{displaymath}$

(3)

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

$\begin{displaymath} y_1[y''_2+P(x)y'_2+Q(x)y_2]-y_2[y''_1+P(x)y'_1+Q(x)y_1]=0 \end{displaymath}$

(4)

$\begin{displaymath} (y_1y''_2-y_2y''_1)+P(y_1y'_2-y'_1y_2)+Q(y_1y_2-y_1y_2)=0 \end{displaymath}$

(5)

$\begin{displaymath} (y_1y''_2-y_2y''_1)+P(y_1y'_2-y'_1y_2)=0. \end{displaymath}$

(6)

Now, use the definition of the Wronskian and take its Derivative,

Plugging

and

into (6) gives

$\begin{displaymath} W'+PW = 0. \end{displaymath}$

(9)

This can be rearranged to yield

$\begin{displaymath} {dW\over W} = -P(x)\,dx \end{displaymath}$

(10)

which can then be directly integrated to

$\begin{displaymath} \ln\left[{W(x)\over W_0}\right]= - \int P(x)\,dx, \end{displaymath}$

(11)

where $\ln x$ is the Natural Logarithm. Exponentiation then yields Abel's identity

$\begin{displaymath} W(x) = W_0e^{-\int P(x)\,dx}, \end{displaymath}$

(12)

where

is a constant of integration.

References

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.