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DIFFERENTIAL EQUATIONS MA133                               CHEAT SHEET


This is a summary sheet for the first half of the module, solving first and second
order ordinary differential equations. We assume conditions for FTC hold. You
should learn all these techniques by heart, and practice, practice, practice!

First Order Differential Equations We consider the main scenarios

     Trivial Case (Section 1.1)

                                               dx
                                                  = f (t)
                                               dt
          By Fundamental Theorem of Calculus simply integrate both sides with
          respect to t
                                  x(t) = f (t) dt

     Linear Non-homogeneous (Sections 1.3/1.4/1.5)

                                        dx
                                           + p(t)x = q(t)
                                        dt

          Multiply both sides by an Integrating Factor P (t) = exp ( p(t)dt) so that

                                     d
                                        (P (t)x(t)) = P (t)q(t)
                                     dt
          Then integrate so that

                             x(t) = P (t)−1        P (s)q(s)ds + AP (t)−1
                                               t

     Separable Equations (Section 1.6)

                                          dx
                                             = f (x)g(t)
                                          dt

          First look for constant solutions, i.e. where f (x) = 0. Then look for non-
          constant solutions (so f (x) never zero) and ”divide both sides by f (x),
          multiply both sides by dt and integrate”.

                                            dx
                                                 =       g(t)dt
                                           f (x)

     Autonomous First Order ODEs (Section 1.9)

                                              dx
                                                 = f (x)
                                              dt

          Look for fixed points x∗ , which satisfy f (x∗ ) = 0, i.e. are points where
          dx
             = 0. A fixed point x∗ is stable if f ′ (x∗ ) < 0 and unstable if f ′ (x∗ ) > 0.
          dt



                                           1
Second Order Ordinary Differential Equations With Constant Coefficients


                                       d2 x    dx
                                            + b + cx = f (t)
                                   a      2
                                       dt      dt

    The solution consists of x(t) = xc (t)+xp (t) where xc (t), the complementary so-
    lution, solves the homogeneous case f (t) = 0 and xp (t), the particular integral,
    gives the f (t).

    The Complementary Solution Solves

                                             d2 x    dx
                                                  + b + cx = 0
                                         a      2
                                             dt      dt
         Find the roots to the auxiliary equations

                                             aλ2 + bλ + c = 0
                        √
                     −b± b2 −4ac
         i.e. λ± =                 then we have
                         2a

           • Real roots k1 , k2 complementary solution is

                                                   Aek1 t + Bek2 t

           • Repeated real root k complementary solution is

                                                   Aekt + Btekt

           • Complex roots p ± iq complementary solution is

                                         ept (A sin(qt) + B cos(qt))

             or
                                               Aept cos(qt − φ)
    The Particular Integral Functions to ”guess”:


                      f (t)                           Try solution xp (t) =
              aekt (k not a root)                             Aekt
                aekt (k a root)                          Atekt or At2 ekt
            a sin(ωt) or a cos(ωt)                   A sin(ωt) + B cos(ωt)
               atn where n ∈ N             P (t) general polynomial of degree n
                     atn ekt          P (t)ekt , P (t) general polynomial of degree n
            n
           t (a sin(ωt) + b cos(ωt)              P1 (t) sin(ωt) + P2 (t) cos(ωt)
                                       where Pi (t) general polynomial of degree n
          ekt (a sin(ωt) + b cos(ωt))             ekt (A sin(ωt) + B cos(ωt))




                                               2

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  • 1. DIFFERENTIAL EQUATIONS MA133 CHEAT SHEET This is a summary sheet for the first half of the module, solving first and second order ordinary differential equations. We assume conditions for FTC hold. You should learn all these techniques by heart, and practice, practice, practice! First Order Differential Equations We consider the main scenarios Trivial Case (Section 1.1) dx = f (t) dt By Fundamental Theorem of Calculus simply integrate both sides with respect to t x(t) = f (t) dt Linear Non-homogeneous (Sections 1.3/1.4/1.5) dx + p(t)x = q(t) dt Multiply both sides by an Integrating Factor P (t) = exp ( p(t)dt) so that d (P (t)x(t)) = P (t)q(t) dt Then integrate so that x(t) = P (t)−1 P (s)q(s)ds + AP (t)−1 t Separable Equations (Section 1.6) dx = f (x)g(t) dt First look for constant solutions, i.e. where f (x) = 0. Then look for non- constant solutions (so f (x) never zero) and ”divide both sides by f (x), multiply both sides by dt and integrate”. dx = g(t)dt f (x) Autonomous First Order ODEs (Section 1.9) dx = f (x) dt Look for fixed points x∗ , which satisfy f (x∗ ) = 0, i.e. are points where dx = 0. A fixed point x∗ is stable if f ′ (x∗ ) < 0 and unstable if f ′ (x∗ ) > 0. dt 1
  • 2. Second Order Ordinary Differential Equations With Constant Coefficients d2 x dx + b + cx = f (t) a 2 dt dt The solution consists of x(t) = xc (t)+xp (t) where xc (t), the complementary so- lution, solves the homogeneous case f (t) = 0 and xp (t), the particular integral, gives the f (t). The Complementary Solution Solves d2 x dx + b + cx = 0 a 2 dt dt Find the roots to the auxiliary equations aλ2 + bλ + c = 0 √ −b± b2 −4ac i.e. λ± = then we have 2a • Real roots k1 , k2 complementary solution is Aek1 t + Bek2 t • Repeated real root k complementary solution is Aekt + Btekt • Complex roots p ± iq complementary solution is ept (A sin(qt) + B cos(qt)) or Aept cos(qt − φ) The Particular Integral Functions to ”guess”: f (t) Try solution xp (t) = aekt (k not a root) Aekt aekt (k a root) Atekt or At2 ekt a sin(ωt) or a cos(ωt) A sin(ωt) + B cos(ωt) atn where n ∈ N P (t) general polynomial of degree n atn ekt P (t)ekt , P (t) general polynomial of degree n n t (a sin(ωt) + b cos(ωt) P1 (t) sin(ωt) + P2 (t) cos(ωt) where Pi (t) general polynomial of degree n ekt (a sin(ωt) + b cos(ωt)) ekt (A sin(ωt) + B cos(ωt)) 2