# 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}

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