Reduction of order

From Conservapedia

Reduction of order is a process used to find the solution of a differential equation $y'' + p (t) y' + q (t) y = 0$ , when Euler substitution methods find only one value for $λ$ . (That is, $\sqrt {b^2-4ac} = 0$ and $\lambda = \frac{-b}{2a}$ .)

The process is carried out in the following manner:

1. $y 1 = e λ t$ . However, we must find $y 2$ . $y 2$ cannot be a multiple of $y 1$ , so we assume that $y 2 = v (t) y 1$ .

2. We apply the product rule to find:

$y 2 = v (t) y 1$

$y 2' = v (t)' y 1 + v (t) y 1'$

$y 2'' = v (t)'' y 1 + 2 v (t)' y 1' + v (t) y 1''$

3. We substitute these expressions into the initial differential equation:

$y 2'' + p (t) y 2' + q (t) y = (v (t)'' y 1 + 2 v (t)' y 1' + v (t) y 1'') + p (t)(v (t)' y 1 + v (t) y 1') + q (t)(v (t) y 1)$

4. We collect the $v (t)''$ , $v (t)'$ , and $v (t)$ terms:

$v (t)''(y 1) + v (t)'(2 y 1' + p (t) y 1) + v (t)(y 1'' + p (t) y 1' + q (t) v (t) y 1)$

5. The terms $(2 y 1' + p (t) y 1)$ and $(y 1'' + p (t) y 1' + q (t) v (t) y 1)$ equal zero in most instances, leading to the conclusion:

$v (t)''(y 1) = 0$

Integrating twice, we yield:

$v (t) = c 1 t + c 2$

6. This is enough to say that $y 2 = v (t) y 1 = t y 1 = t e λ t$

7. The solution is then $y = c 1 y 1 + c 2 y 2 = e λ t (c 1 + c 2 t)$

Reduction of order

From Conservapedia

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