3.3 Complex Roots and Characteristic Equation

Charecteristic equations

• when the determinent ($b^2-4ac$) is equal to 0 we know that there will only be 1 r value

  • then we know for a fact that $e^{rt}$ is a solution!

  • how do we find the second independent solution

    • make an educated gues that the second solution is
      • $y_2(t)=v(t)y_1(t)=v(t)e^{-\left(\frac{b}{2a}\right)t}$ for some unknown v(t)

  • so then we put our educated guess into the seconed order equation

    • $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0⟹y_2^{\prime }(t)=v^{\prime }(t)e^{-\frac{b}{2a}t}-\frac{b}{2a}v(t)e^{-\frac{b}{2a}t}$
    • $y_2^{\prime \prime }=v^{\prime \prime }(t)e^{-\frac{b}{2a}t}-\frac{b}{2a}{v}^{\prime }(t)e^{-\frac{b}{2a}t}-\frac{b}{2a}{v}^{\prime }(t)$

  • This becomes

    • $y_2=K_{1}t{e}^{-\left(\frac{b}{2a}\right)t}+K_2{e}^{-\frac{b}{2a}t}$
    • $y_1(t)=e^{-\frac{b}{2a}t}$
    • so that way we can pick $k_1=1$ $k_2=0$ so that $t{e}^{-\frac{b}{2a}t}$ is our second solution

    • So our solutions are

      • $y(t)=c_1{e}^{-\frac{b}{a}t}+c_{2}t{e}^{-\frac{b}{2a}t}$

Hyperbolic functions

• cosh(x) and sinh(x)

  • Definition

  • $\cosh x=\frac{e^x+e^{-x}}{2}$
  • $\sinh x=\frac{e^x-e^{-x}}{2}$

  • These are called hyperbolic because on a hyperbola $x^2-y^2=1$ we can write any point of it as

    • $\cosh t,\sinh t$

• When do we use this?

  • These are alternate forms of writing exponential solution

    • useful when ODEs

  • consider the simple 3.1 Second-order homogeneous Differential Equations

    • $y^{\prime \prime }-a^{2}y=0$
      • the characteristic equation is $r^2-a^2=0$ with roots $r=\pm a$

    • Then we can rewrite instead of using exponitel we can use

      • $y(x)=\sinh (ax)-\cosh (ax)$

• Example

  • solve the ivp $y^{\prime \prime }+2y^{\prime }+5y=0,y(0)=0,y^{\prime }(0)=1$

$$\begin{gathered} \begin{array}{r}\text{a) find the roots of the characteristic equation:}\\ r^2+2r+5=0⟹-1\pm \frac{i}{2}⟹\lambda =-1,\mu =\frac{1}{2}\\ \\ \text{b) find general equation}\\ y_1(t)=e^{\lambda t}\sin \mu t=e^{-t}\cos \left(\frac{1}{2}t\right)\\ y_2(t)=e^{\lambda t}\sin \mu t=e^{-t}\sin \left(\frac{1}{2}t \\ \right)\\ y(t)=c_1{e}^{-t}\cos \frac{t}{2}+c_2{e}^{-t}\sin \frac{t}{2}\\ \\ \text{c) apply initial conditions}\\ y(0)=0\dots \\ y(t)=e^{-t}\sin \frac{t}{2}\end{array} \end{gathered}$$

• notes about exponential and trig functions being multiplied

  • when they are being multiplied we are limited to the exponential as the function mill bounce around like its a ceiling