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 nonhomogeneous equation.
y''-25y= 4; y1=e^-5x

a. y2(x) = ?
b. yp(x) = ?

Answer 1

Answer:

a)  y₂ (x) = e ⁵ˣ  

Complementary function

               $y_{C} = C_{1} {e^{-5x} } + C_{2} {e^{5x} }$

b) particular integral

$P.I = y_{p} = \frac{-4}{25}$

Step-by-step explanation:

step(i):-

Given differential equation y''-25y= 4

operator form

             ⇒    D²y - 25 y =4

            ⇒     (D² - 25) y =4

       This is the form of f(D)y = ∝(x)

where f(m) = D² - 25     and ∝(x) =4

The auxiliary equation A(m) =0

                         ⇒ m² - 25 =0

                          m² - 5²  =0

                      ⇒ (m+5)(m-5) =0

                     ⇒ m =-5 , 5

Complementary function

               $y_{C} = C_{1} {e^{-5x} } + C_{2} {e^{5x} }$

This is form of

             $y_{C} = C_{1} y_{1} (x) + C_{2} y_{2} (x)$

where y₁ (x) = e⁻⁵ˣ   and  y₂ (x) = e ⁵ˣ  

Step(ii):-

Particular integral:-

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

$P.I = y_{p} = \frac{1}{D^{2} -25} 4$

      =  $= \frac{1}{D^{2} -25} 4e^{0x}$

put D = 0

The particular integral

$y_{p} = \frac{1}{ -25} 4$

$P.I = y_{p} = \frac{-4}{25}$

Conclusion:-

General solution of given differential equation

$y = y_{C} +y_{P}$

$y = C_{1} {e^{-5x} } + C_{2} {e^{5x} } -\frac{4}{25}$