Non-homogeneous equations (Sect. 3.6).
              ◮ We study: y′′ +p(t)y′ +q(t)y = f(t).
              ◮ Method of variation of parameters.
              ◮ Using the method in an example.
              ◮ The proof of the variation of parameter method.
              ◮ Using the method in another example.
         Method of variation of parameters.
             Remarks:
              ◮ This is a general method to find solutions to equations having
                variable coefficients and non-homogeneous with a continuous
                but otherwise arbitrary source function,
                             y′′ + p(t)y′ + q(t)y = f (t).
              ◮ The variation of parameter method can be applied to more
                general equations than the undetermined coefficients method.
              ◮ The variation of parameter method usually takes more time to
                implement than the simpler method of undetermined
                coefficients.
            Method of variation of parameters.
                 Theorem (Variation of parameters)
                 Let p, q, f : (t1,t2) → R be continuous functions, let y1,
                 y2 : (t1, t2) → R be linearly independent solutions to the
                 homogeneous equation
                                     y′′ + p(t)y′ + q(t)y = 0,
                 and let W     be the Wronskian of y and y . If the functions u
                           y y                       1       2                  1
                            1 2
                 and u are defined by
                      2        Z                              Z
                       u (t) =    −y2(t)f(t) dt,      u (t) =    y1(t)f (t) dt,
                        1           W (t)              2         W (t)
                                       y y                          y y
                                        1 2                          1 2
                 then the function y = u y +u y is a particular solution to the
                                    p     1 1    2 2
                 non-homogeneous equation
                                    y′′ + p(t)y′ + q(t)y = f (t).
            Non-homogeneous equations (Sect. 3.6).
                   ◮ We study: y′′ +p(t)y′ +q(t)y = f(t).
                   ◮ Method of variation of parameters.
                   ◮ Using the method in an example.
                   ◮ The proof of the variation of parameter method.
                   ◮ Using the method in another example.
            Using the method in an example.
                 Example
                 Find the general solution of the inhomogeneous equation
                                        y′′ − 5y′ + 6y = 2et.
                 Solution:
                 First: Find fundamental solutions to the homogeneous equation.
                 The characteristic equation is                           (
                                                1     √        
            r =3,
                  r2 −5r +6 = 0       ⇒ r= 5± 25−24                  ⇒       1
                                                2                            r =2.
                                                                             2
                 Hence, y (t) = e3t and y (t) = e2t. Compute their Wronskian,
                          1               2
                   W (t)=(e3t)(2e2t)−(3e3t)(e2t)            ⇒ W (t)=−e5t.
                     y y                                           y y
                      1 2                                           1 2
                 Second: We compute the functions u and u . By definition,
                                                       1      2
                                   u′ = − y2f ,        u′ = y1f .
                                    1      W            2   W
                                             y y              y y
                                              1 2              1 2
            Using the method in an example.
                 Example
                 Find the general solution of the inhomogeneous equation
                                        y′′ − 5y′ + 6y = 2et.
                 Solution: Recall: y (t) = e3t, y (t) = e2t, W    (t) = −e5t, and
                                    1           2             y y
                                                               1 2
                                   u′ = − y2f ,        u′ = y1f .
                                    1      W            2   W
                                             y y              y y
                                              1 2              1 2
                   u′ = −e2t(2et)(−e−5t)      ⇒ u′=2e−2t         ⇒ u =−e−2t,
                    1                               1                   1
                    u′ = e3t(2et)(−e−5t)     ⇒ u′=−2e−t ⇒ u =2e−t.
                     2                              2                    2
                 Third: The particular solution is
                          y =(−e−2t)(e3t)+(2e−t)(e2t)         ⇒ y =et.
                           p                                         p
                 The general solution is y(t) = c e3t + c e2t + et, c ,c ∈ R.     ⊳
                                                 1       2           1  2
            Non-homogeneous equations (Sect. 3.6).
                  ◮ We study: y′′ +p(t)y′ +q(t)y = f(t).
                  ◮ Method of variation of parameters.
                  ◮ Using the method in an example.
                  ◮ The proof of the variation of parameter method.
                  ◮ Using the method in another example.
            The proof of the variation of parameter method.
                Proof: Denote L(y) = y′′ +p(t)y′ +q(t)y.
                Weneed to find yp solution of L(yp) = f.
                Weknow y and y solutions of L(y ) = 0 and L(y ) = 0.
                           1      2               1             2
                Idea: The reduction of order method: Find y proposing y = uy .
                                                           2            2     1
                First idea: Propose that y is given by y = u y + u y .
                                         p             p    1 1    2 2
                Wehope that the equation for u and u will be simpler than the
                                                1      2
                original equation for yp, since y1 and y2 are solutions to the
                homogeneous equation. Compute:
                                  y′ = u′y +u y′ +u′y +u y′,
                                   p    1 1    1 1   2 2    2 2
                         y′′ = u′′y + 2u′y′ + u y′′ + u′′y + 2u′y′ + u y′′.
                          p    1  1     1 1   1 1     2 2     2 2   2 2