Abel's Identity for Linear Homogenous Second Order Differential Eqs.

Abel's Identity for Linear Homogenous Second Order Differential Equations

Recall from the Wronskian Determinants and Linear Homogenous Differential Equations page that if we have a linear homogenous second order differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ and $y = y_1(t)$ and $y = y_2(t)$ are solution to this differential equation, then the Wronskian of $y_1$ and $y_2$ is defined as:

(1)
\begin{align} W(y_1, y_2) = \begin{vmatrix} y_1(t) & y_2(t)\\ y_1'(t) & y_2'(t) \end{vmatrix} \end{align}

The following theorem gives us an alternative form for the Wronskian between the solutions $y_1$ and $y_2$ of this differential equation.

 Theorem 1 (Abel's Identity for Linear Homogenous Second Order Differential Equations): Let $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = 0$ be a second order linear homogenous differential equation where $p$ and $q$ are continuous on an open interval $I$ such that $t_0 \in I$. Then the Wronskian of $y_1$ and $y_2$ at some $t$ is given by $W(y_1, y_2) = C e^{- \int p(t) \: dt}$ where $C$ is some constant dependent on $y_1$ and $y_2$.
• Proof: Let $y = y_1(t)$ and $y = y_2(t)$ be solutions to this differential equation. Then we have that:
(2)
\begin{align} \quad \frac{d^2 y_1}{dt^2} + p(t) \frac{d y_1}{dt} + q(t) y_1 = 0 \: (*) \quad \mathrm{and} \quad \frac{d^2 y_2}{dt^2} + p(t) \frac{d y_2}{dt} + q(t) y_2 = 0 \: (**) \end{align}
• Now take the first equation, $(*)$, and multiply both sides by $-y_2(t)$ to get:
(3)
\begin{align} \quad -y_2(t)\frac{d^2 y_1}{dt^2} -y_2(t)p(t)\frac{d y_1}{dt} - y_2(t)q(t) y_1 = 0 \end{align}
• Now take the second equation, $(**)$ and multiply both sides by $y_1(t)$ to get:
(4)
\begin{align} \quad y_1(t) \frac{d^2 y_2}{dt^2} + y_1(t) p(t) \frac{d y_2}{dt} + y_1(t) q(t) y_2 = 0 \end{align}
• We will then add the last two equations together and so:
(5)
\begin{align} \quad 0 = \left [y_1(t) \frac{d^2 y_2}{dt^2} + y_1(t) p(t) \frac{d y_2}{dt} + y_1(t) q(t) y_2 \right ] + \left [-y_2(t)\frac{d^2 y_1}{dt^2} - y_2(t) p(t) \frac{d y_1}{dt} - y_2(t)q(t) y_1 \right ] \\ = \left ( y_1(t) \frac{d^2 y_2}{dt^2} - y_2 \frac{d^2 y_1}{dt^2} \right ) + p(t) \left ( y_1(t) \frac{d y_2}{dt} - y_2(t) \frac{d y_1}{dt} \right ) + q(t) \underbrace{\left ( y_1(t) y_2(t) - y_1(t) y_2(t) \right )}_{= 0} \\ = \left ( y_1(t) \frac{d^2 y_2}{dt^2} - y_2 \frac{d^2 y_1}{dt^2} \right ) + p(t) \left ( y_1(t) \frac{d y_2}{dt} - y_2(t) \frac{d y_1}{dt} \right ) \end{align}
• Now the Wronskian of $y_1$ and $y_2$ is given by:
(6)
\begin{align} \quad W(t) = W(y_1, y_2) = \begin{vmatrix} y_1(t) & y_2(t)\\ y_1'(t) & y_2'(t)\end{vmatrix} = \begin{vmatrix} y_1(t) & y_2(t)\\ \frac{d y_1}{dt} & \frac{d y_2}{dt} \end{vmatrix} = y_1(t) \frac{d y_2}{dt} - y_2(t) \frac{d y_1}{dt} \end{align}
• Note also that:
(7)
\begin{align} \quad W'(t) = y_1'(t)\frac{d y_2}{dt} + y_1(t) \frac{d^2 y_2}{dt^2} - y_2'(t) \frac{d y_1}{dt} - y_2 \frac{d^2 y_1}{dt^2} = y_1(t) \frac{d^2 y_2}{dt^2} - y_2 \frac{d^2 y_1}{dt^2} + \underbrace{y_1(t) y_2'(t) - y_1(t)y_2'(t)}_{=0} = y_1(t) \frac{d^2 y_2}{dt^2} - y_2 \frac{d^2 y_1}{dt^2} \end{align}
• From this, we can now rewrite the equation from earlier in the form $W'(t) + p(t) W(t) = 0$. We can solve this differential equation by using the method of integrating factors. Let $\mu (t) = e^{\int p(t) \: dt}$ and so:
(8)
\begin{align} W'(t) + p(t) W(t) = 0 \\ \mu (t) W'(t) + \mu (t) p(t) W(t) = 0 \\ \frac{d}{dt} \left ( W(t) \mu (t) \right ) = 0 \\ \int \frac{d}{dt} \left ( W(t) \mu (t) \right ) \: dt = \int 0 \: dt \\ W(t) \mu(t) = C \\ W(t) = \frac{C}{\mu (t)} \\ W(t) = \frac{C}{e^{\int p(t) \: dt}} \\ W(t) = Ce^{- \int p(t) \: dt} \quad \blacksquare \end{align}