3.1 Second-order homogeneous Differential Equations

Second-Order ODES

• General form:

  • $\frac{d^{2}y}{d{t}^2}=f\left(t,y(t),\frac{dy}{dt}(t)\right)$

• Linear vs non linear

  • a second-order ODE is linear if it can be written as
    • $y,,+p(t)y^{\prime }+q(t)y=g(t)$
  • other wise, the ode is non-linear

• Homogeneous vs non-homogeneous:

  • A second order Linear ODE is called homogeneous if the RHS function $g(t)=0$, or $G(t)=0$

  • other wise, it is called non homogeneous
    • to solve these equations we just think it is homogenous and solve that and then bring back that solution to the origanal function

• Initial conditions

  • to obtian a unique solution, 2 initial conditions must be imposed:

Boundry value problems

• Definition

  • insteaf of initial conditions, you can determine a unque solution by imposing two conditions on y only at the endpoints of an interval [t0, t1]

Example

find the general solution

• $y^{\prime \prime }+5y^{\prime }+6y=0$
  • recall: the ode y’=ry has solution $y(t)=c{e}^{rt}$
    • idea: try out the same exponential form of solution y(t)=$e^{rt}$

$$\begin{array}{r}y(t)=e^{rt},y^{\prime }(t)={\mathrm{re}}^{rt},y^{\prime \prime }(t)=r^2{e}^{rt}\\ \text{ plugging this into the eq ->}{r}^2{e}^{rt}+5(r{e}^{rt})+6e^{rt}=0\\ ⟹e^{rt}(r^2+5r+6)=0\\ ⟹(r+2)(r+3)=0\to r=-2,y(t)=e^{-2t}\to r=-,y(t)=e^{-3t}\\ y(t)=c_1{e}^{-2t}+c_2{e}^{-3t}\\ \text{plugging it into the equation again}\\ ⟹(4c_1{e}^{-2t}+9c_2{e}^{-3})+5(-2c_1{e}^{-2t}-3c_2{e}^{-3t})+6(c_1{e}^{-2t}+c_2{e}^{-3t})=0\\ y(t)=c_1{e}^{-2t}+c_2{e}^{-3t}\end{array}$$