Question

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation and a particular solution yp(x) of the given non-homogeneous equation.
y'' − 5y' + 4y = x; y1 = ex
Find yp(x).

Answer 1

Answer:

The particular integral of given differential equation

                 $y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$

General solution of given differential equation

     $y = y_{c} + y_{p}$

 $Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$

Step-by-step explanation:

Step(i):-

Given Differential equation  y'' − 5 y' + 4 y = x

Given equation in operator form

        D²y - 5 Dy +  4 y = x

⇒     ( D² - 5 D +  4 ) y =x

⇒    f(D) y = Q

where  f(D) = D² - 5 D +  4 and Q(x) = x

The auxiliary equation  f(m) =0

          m²-5 m + 4 =0

         m² - 4 m - m + 4 =0

        m ( m -4 ) -1 ( m-4) =0

         (m - 1) =0   and ( m-4) =0

        m = 1 and m =4

The complementary function

$Y_{c} = C_{1} e^{x} + C_{2} e^{4x}$

Step(ii):-

particular integral

Particular integral

    $y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2} - 5 D + 4} X$

taking common '4'

                         $= \frac{1}{4(1 + (\frac{D^{2} - 5 D}{4} ))} X$

                       

                          $=\frac{1}{4} (1 + (\frac{D^{2} -5D}{4})^{-1} )} X$

applying binomial expression

     ( 1 + x )⁻¹    = 1 - x + x² - x³ +.....      

                         $=\frac{1}{4} (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X$

Now simplifying and we will use notation D = $\frac{dy}{dx}$

                       $=\frac{1}{4} (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}$

Higher degree terms are neglected

                    $=\frac{1}{4} (x - (\frac{ -5 D}{4}) x)$

The particular integral of given differential equation

                 $y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))$

Final answer:-

         General solution of given differential equation

     $y = y_{c} + y_{p}$

 $Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))$