Respuesta :
The question is :
2x²y'' + 5xy' + y = x² - x;
y = c1x^(1/2) + c2x^(-1) + 1/15(x^2) - 1/6(x), (0,infinity)
The functions (x^-1/2) and (x^-1) satisfy the differential equation and are linearly independent since W(x^-1/2, x^-1)= ____?____ for 0
Answer:
The functions x^(-1/2) and x^(-1) are linearly independent since their wronskian is (-1/2)x^(-5/2) ≠ 0.
Step-by-step explanation:
Suppose the functions x^(-1/2) and x^(-1) satisfy the differential equation 2x²y'' + 5xy' + y = x² - x;
and are linearly independent, then their wronskian is not zero.
Wronskian of y1 and y2 is given as
W(y1, y2) = y1y2' - y1'y2
Let y1 = x^(-1/2)
y1' = (-1/2)x^(-3/2)
Let y2 = x^(-1)
y2' = -x^(-2)
W(y1, y2) =
x^(-1/2)(-x^(-2)) - (-1/2)x^(-3/2)x^(-1)
= -x^(-5/2) + (1/2)(x^(-5/2)
= (-1/2)x^(-5/2)
So, W(y1, y2) = (-1/2)x^(-5/2) ≠ 0
Which means the functions are linearly independent.
The functions [tex]\rm (x^\frac{-1}{2})[/tex] and [tex]\rm (x^{-1})[/tex] satisfy the differential equation and are linearly independent since [tex]W(x^{-1/2}, x^{-1})[/tex] = [tex](-x^{5/2} )+ \dfrac{1}{2}.(x^{5/2})[/tex] .
Given that,
The given two-parameter family of functions is the general solution of the non-homogeneous differential equation on the indicated interval.
[tex]\rm 2x^2y^n+5xy'+y = x^2-x[/tex]
[tex]\rm y = c_1x^{\frac{-1}{2} }+c_2{-x} \dfrac{1}{15}x^2=\dfrac{1}{5}x ,[/tex]
We have to determine,
The functions [tex]\rm (x^\frac{-1}{2})[/tex] and [tex]\rm (x^{-1})[/tex] satisfy the differential equation and are linearly independent since [tex]W(x^{-1/2}, x^{-1})[/tex] for 0?
According to the question,
The functions [tex]\rm (x^\frac{-1}{2})[/tex] and [tex]\rm (x^{-1})[/tex] satisfy the differential equation,
[tex]\rm 2x^2y^n+5xy'+y = x^2-x[/tex]
And are linearly independent, then their differentiation is not zero.
The differential equation is given by,
[tex]\rm Y(y_1, y_2) = y_1.y_2'- y_2.y_1'[/tex]
The value [tex]\rm y_1'[/tex] is,
[tex]\rm y_1 = x^{(-1/2})\\\\ y_1' = \dfrac{-1}{2} x^{(-3/2)}[/tex]
And value of [tex]\rm y_2'[/tex]
[tex]\rm y_2 = x^{(-1)}\\\\y_2' = -x^{(-2)}[/tex]
Therefore,
[tex]\rm Y(y_1, y_2) = (x^{-1/2}.(-x)^{-2}-(-\dfrac{1}{2}.x^{-3/2}).(x^{-1})\\\\\rm Y(y_1, y_2) = (-x^{5/2} )+ \dfrac{1}{2}.(x^{5/2})\neq 0\\\\[/tex]
Hence, The value of the function is not equal to zero then the function is linearly independent.
For more details refer to the link given below.
https://brainly.com/question/18510715
