Respuesta :
Answer:
a) y₂ (x) = e ⁵ˣ
Complementary function
[tex]y_{C} = C_{1} {e^{-5x} } + C_{2} {e^{5x} }[/tex]
b) particular integral
[tex]P.I = y_{p} = \frac{-4}{25}[/tex]
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
[tex]y_{C} = C_{1} {e^{-5x} } + C_{2} {e^{5x} }[/tex]
This is form of
[tex]y_{C} = C_{1} y_{1} (x) + C_{2} y_{2} (x)[/tex]
where y₁ (x) = e⁻⁵ˣ and y₂ (x) = e ⁵ˣ
Step(ii):-
Particular integral:-
[tex]P.I = y_{p} = \frac{1}{f(D)} \alpha (x)[/tex]
[tex]P.I = y_{p} = \frac{1}{D^{2} -25} 4[/tex]
= [tex]= \frac{1}{D^{2} -25} 4e^{0x}[/tex]
put D = 0
The particular integral
[tex]y_{p} = \frac{1}{ -25} 4[/tex]
[tex]P.I = y_{p} = \frac{-4}{25}[/tex]
Conclusion:-
General solution of given differential equation
[tex]y = y_{C} +y_{P}[/tex]
[tex]y = C_{1} {e^{-5x} } + C_{2} {e^{5x} } -\frac{4}{25}[/tex]
