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

a) y₂ (x) = e ⁵ˣ

Complementary function

`y_(C) = C_(1) {e^(-5x) } + C_(2) {e^(5x) }`

b) particular integral

`P.I = y_(p) = (-4)/(25)`

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) = (1)/(f(D)) \alpha (x)`

`P.I = y_(p) = (1)/(D^(2) -25) 4`

=
`= (1)/(D^(2) -25) 4e^(0x)`

put D = 0

The particular integral

`y_(p) = (1)/( -25) 4`

`P.I = y_(p) = (-4)/(25)`

Conclusion:-

General solution of given differential equation

`y = y_(C) +y_(P)`

`y = C_(1) {e^(-5x) } + C_(2) {e^(5x) } -(4)/(25)`