Table of Contents
Differential Equations
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Topic PlaylistThis method is useful for finding a second homogenous solution to a linear second order ODE when one homogeneous solution is known. This already happened for the $ay''+by'+cy = 0$ Repeated Real Root case. When the only known solution was $e^{rx}$ where $r$ was the repeated real root of the characteristic equation $ar^{2}+br+c = 0$, the second solution $xe^{rx}$ can be obtained using reduction of order.
Solution VideoFind the second homogeneous solution to the ODE when you are given only one homogeneous solution.
\begin{equation} y'' - 8y' + 16y = 0 \qquad y_{1}(x) = e^{4x} \end{equation} Do this by guessing the second homogeneous solution is of the form $y_{2}(x) = v(x)e^{4x}$ where $e^{4x}$ comes from the first homogeneous solution. Solution VideoFind the general solution to the ODE.
\begin{equation} x^{2}y'' - 3xy' + 3y = 0 \qquad y_{1}=x \end{equation} Solution VideoFind the general solution to the ODE.
\begin{equation} x^{2}\big(\ln|x|\big)^{2}y''-2x\ln|x|y'+\big(2+\ln|x|\big)y=0 \qquad y_{1}=\ln|x| \end{equation} Solution VideoFind the general solution to the ODE.
\begin{equation} (x-1)y''-xy'+y=0 \qquad y_{1}=e^{x} \end{equation} Explanation VideoReduction of order uses a guess $y_{2}(x) = v(x)y_{1}(x)$ which results in a new ODE for the function $v(x)$ with $v''(x)$ and $v'(x)$ terms, but no $v(x)$ term. There is no $v(x)$ term because the half of the product rule for the guess $v(x)y_{1}(x)$ that takes the derivative of $y_{1}(x)$ acts like the first homogeneous solution when plugged into the ODE.
\begin{equation} y''+p(x)y'+q(x)y=0 \qquad y_{2}(x)=v(x)y_{1}(x) \end{equation} \begin{equation} y_{2}''+p(x)y_{2}'+q(x)y_{2}=(vy_{1})''+p(x)(vy_{1})'+q(x)vy_{1}=0 \end{equation} \begin{equation} v''y_{1}+2v'y_{1}'+vy_{1}''+p(x)(v'y_{1}+vy_{1}')+q(x)vy_{1}=0 \end{equation} \begin{equation} y_{1}v''+\big(2y_{1}'+p(x)y_{1}\big)v'+\big(y_{1}''+p(x)y_{1}'+q(x)y_{1}\big)v=0 \end{equation} \begin{equation} y_{1}v''+\big(2y_{1}'+p(x)y_{1}\big)v'+0v=0 \end{equation} The $\big(y_{1}''+p(x)y_{1}'+q(x)y_{1}\big)v$ was replaced by $0v$ because $y_{1}$ is a homogeneous solution to the original ODE, and the original ODE is exactly what is in the parentheses, so all the $v(x)$ terms disappear. \begin{equation} y_{1}v''+\big(2y_{1}'+p(x)y_{1}\big)v'=0 \end{equation} Even though this is a second order ODE, we can use first order solution methods because there is no $v(x)$ term. To make it look like a first order ODE, we can replace $v'(x)=w(x)$ so it's not as confusing. \begin{equation} w(x) = v'(x) \qquad y_{1}w'+\big(2y_{1}'+p(x)y_{1}\big)w=0 \end{equation} Since $p(x)$ and $y_{1}(x)$ are known, we can solve for $w(x)$, which can be used to find $v(x)$, which can be used to find the $y_{2}(x)$ guess we started with.