Reduction of order

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 $\lambda$. (That is, ${\sqrt {b^{2}-4ac}}=0$ and $\lambda ={\frac {-b}{2a}}$.)

The process is carried out in the following manner:

1. $y_{1}=e^{\lambda t}$. 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:

$y_{2}=v(t)y_{1}$

$y_{2}'=v(t)'y_{1}+v(t)y_{1}'$

$y_{2}''=v(t)''y_{1}+2v(t)'y_{1}'+v(t)y_{1}''$

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

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

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

$v(t)''(y_{1})+v(t)'(2y_{1}'+p(t)y_{1})+v(t)(y_{1}''+p(t)y_{1}'+q(t)v(t)y_{1})$

5. The terms $(2y_{1}'+p(t)y_{1})$ 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 $y_{2}=v(t)y_{1}=ty_{1}=te^{\lambda t}$

7. The solution is then $y=c_{1}y_{1}+c_{2}y_{2}=e^{\lambda t}(c_{1}+c_{2}t)$