Question

Write the given differential equation in the formL(y) = g(x),where L is a linear differential operator with constant coefficients. If possible, factor L. (Use D for the differential operator.)y'' − 3y' − 28y = x − 2

Answer 1

Answer:

L(y) = g(x)

$y = C_{1} e^{-4 x} + C_{2} e^{7 x} + \frac{1}{(-28)}(( x + (\frac{ -3 (1) }{28} )) + \frac{1}{14}$

Step-by-step explanation:

Step(i):-

Given differential equation

  y'' − 3 y' − 28 y = x − 2

Operator form ( D² - 3 D - 28 )y = x -2

                         f( D ) y = Ф(x)

Auxiliary equation

                    f( m ) = m² - 3 m - 28

          ⇒  m² - 3 m - 28 =0

         ⇒  m² - 7 m +4 m - 28 =0

         ⇒ m ( m -7 ) + 4 ( m -7) =0

         ⇒ (m + 4)( m -7) =0

       ⇒ m = -4 and m =7

Complementary function

               $y_{c} = C_{1} e^{-4 x} + C_{2} e^{7 x}$

Step(ii):-

Particular integral

$P.I = y_{p} = \frac{1}{f(D)} (x-2)$

             $= \frac{1}{D^{2}- 3 D - 28 } (x)-2\frac{1}{D^{2} - 3 D - 28} e^{0 x} )$

           $= \frac{1}{(-28)(1 - (\frac{D^{2}-3 D) }{-28} )} (x) + \frac{1}{0-0-28} X 2$

           $= \frac{1}{(-28)}( 1 + (\frac{D^{2} -3 D) }{28} )^{-1} (x) + \frac{1}{14}$

( 1 + x )⁻¹ = 1 - x + x² - x³ +.....

  we use notation $D = \frac{dy}{dx}$

         $= \frac{1}{(-28)}( 1 + (\frac{D^{2} -3 D) }{28} )+ (\frac{D^{2} -3 D) }{28})^{2} +.....) (x) + \frac{1}{14}$

Higher degree terms will be neglected

        D(x) =1

        D²(x) =0

      $= \frac{1}{(-28)}(( x + (\frac{ -3 (1) }{28} )) + \frac{1}{14}$

  The particular integral

  $y_{p} = \frac{1}{(-28)}(( x + (\frac{ -3 (1) }{28} )) + \frac{1}{14}$

Conclusion:-

General solution of given differential equation

$y = y_{c} + y_{p}$

$y = C_{1} e^{-4 x} + C_{2} e^{7 x} + \frac{1}{(-28)}(( x + (\frac{ -3 (1) }{28} )) + \frac{1}{14}$