Table of Contents
Differential Equations
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Topic PlaylistSometimes the solutions to differential equations were guessed by mathematicians. There is no derivation. For the ODE $ay''+by'+cy=0$ someone guessed a solution $y(x)=e^{rx}$ then tried to figure out what value $r$ needed to be to solve the differential equation. The result of plugging in $e^{rx}$ for $y(x)$ is $e^{rx}(ar^{2}+br+c)=0$. The polynomial $ar^{2}+br+c$ is called the characteristic polynomial and solving for the roots gives the values of $r$ to solve the ODE.
Explanation VideoOne of three cases for the characteristic polynomial is imaginary roots.
\begin{equation} ay''+by'+cy=0 \qquad y(x)=e^{rx} \end{equation} \begin{equation} e^{rx}(ar^{2}+br+c)=0 \qquad ae^{rx}\big(r-(\lambda+\mu i)\big)\big(r-(\lambda-\mu i)\big)=0 \qquad i = \sqrt{-1} \end{equation}Even though the roots are imaginary, exponentials with imaginary powers are valid solutions.
\begin{equation} y_{1}(x)=e^{(\lambda+\mu i)x} \qquad y_{2}(x)=e^{(\lambda-\mu i)x} \end{equation}The general solution is,
\begin{equation} y(x)=c_{1}e^{(\lambda+\mu i)x}+c_{2}e^{(\lambda-\mu i)x} \end{equation}However, these are not convenient to work with, especially when solving for a real initial condition. An equivalent solution is,
\begin{equation} y(x)=c_{1}e^{\lambda x}\cos(\mu x)+c_{2}e^{\lambda x}\sin(\mu x) \end{equation} Solution VideoFind the general solution to the ODE.
\begin{equation} y'' + y = 0 \end{equation} Solution VideoFind the general solution to the ODE.
\begin{equation} y'' + 4y' + 13y = 0 \end{equation} Solution VideoSolve the initial value problem.
\begin{equation} y'' + 4y = 0 \qquad y(0) = -2 \qquad y'(0)=5 \end{equation} Solution VideoSolve the initial value problem.
\begin{equation} y'' + 2y' + 2y = 0 \qquad y(0) = 1 \qquad y'(0)= 2 \end{equation} Solution VideoSolve the IVP for a mass spring damper system.
\begin{equation} z'' + 4z' + 8z = 0 \qquad z(0) = -3 \qquad z'(0)= 2 \end{equation} Explanation VideoThe method relies on $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ and combining the constants $c_{1}$ and $c_{2}$ to absorb everything imaginary.
\begin{equation} y(x)=c_{1}e^{(\lambda+\mu i)x}+c_{2}e^{(\lambda-\mu i)x} \end{equation}