Types of ODES

Key take aways

• Every firstorder linear ODE $y^{\prime }=p(t)y=g(t)$ is solvable

• the solution requires TWO integration steps

Summary of the method of integrating factors

• 1. put the ODE in “standard form”

  • $y^{\prime }+p(t)y=g(t)$

• Calculate the integrating factor

  • $u(t)=e^{\int p(t)dt}$

    • for the integrating factor $u(t)$ we can ignore the constant of integration

• Calculate the integral $\int u(t)g(t)dt$ and use the formula

  • $y(t)=\frac{1}{u(t)}\left(\int u(t)g(t)dt\right)$

    • where for the indefinite integral, we need the constant of integration to get the general solution

Note on integral notation:

• ${\int }_{t_0}^{t}u(s)ds$
  • a definite integral
  • t is independent varible
  • s is a dummy varible of integration
• $\int u(s)ds$
  • indefinite integralgenarl
  • a family of functions in s

Ex:

• Find the genarl solution for $y^{\prime }-2y=t^2{e}^{2t}$

  • write ode in standard form:
    • $y^{\prime }-2y=t^2{e}^{2t}$ (left hand =p(t), right hand = (g(t)))
  • calculate IF:
    • $u(t)=e^{\int p(t)dt}⟹u(t)=e^{-2t}$
  • apply formula on the last slide (BUT ONLY DO THIS ONCE)
    • $\frac{d}{dt}(uy)=u{t}^2{e}^{2t}$
      • ⇒ $uy=\int u{t}^2{e}^{2t}dt⟹e^{2t}\left(\frac{t^3}{3}+k\right)$
  • important: always check by subsituting into the ODE
    • $y=\frac{e^{2t}{t}^3}{3}+k{e}^{2t}$