Respuesta :
Answer:
The particular integral of given differential equation
[tex]y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))[/tex]
General solution of given differential equation
[tex]y = y_{c} + y_{p}[/tex]
[tex]Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))[/tex]
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
[tex]Y_{c} = C_{1} e^{x} + C_{2} e^{4x}[/tex]
Step(ii):-
particular integral
Particular integral
[tex]y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2} - 5 D + 4} X[/tex]
taking common '4'
[tex]= \frac{1}{4(1 + (\frac{D^{2} - 5 D}{4} ))} X[/tex]
[tex]=\frac{1}{4} (1 + (\frac{D^{2} -5D}{4})^{-1} )} X[/tex]
applying binomial expression
( 1 + x )⁻¹ = 1 - x + x² - x³ +.....
[tex]=\frac{1}{4} (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X[/tex]
Now simplifying and we will use notation D = [tex]\frac{dy}{dx}[/tex]
[tex]=\frac{1}{4} (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}[/tex]
Higher degree terms are neglected
[tex]=\frac{1}{4} (x - (\frac{ -5 D}{4}) x)[/tex]
The particular integral of given differential equation
[tex]y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))[/tex]
Final answer:-
General solution of given differential equation
[tex]y = y_{c} + y_{p}[/tex]
[tex]Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))[/tex]
