This question is a part of differential equation.
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The complementary function of the given differential equation is the general solution to the homogeneous version of the equation, which is obtained by setting the equation equal to zero.
The differential equation is:
(D^8 + 6D^6 - 32D^2)y = 0
The complementary function, denoted as y_c, is the general solution to the homogeneous equation:
(D^8 + 6D^6 - 32D^2)y_c = 0
To find the complementary function, we assume a solution of the form:
y_c = e^(rt)
Substituting this into the homogeneous equation, we get:
(r^8 + 6r^6 - 32r^2)e^(rt) = 0
Since e^(rt) is never zero, we can cancel it from both sides of the equation:
r^8 + 6r^6 - 32r^2 = 0
This is a polynomial equation in r, and by solving it, we can find the values of r. The complementary function will then be the linear combination of the solutions e^(rt) with the corresponding values of r.
Remember that solving the polynomial equation is a complex task, and the complementary function will depend on the roots (values of r) obtained from the polynomial equation.
The differential equation is:
(D^8 + 6D^6 - 32D^2)y = 0
The complementary function, denoted as y_c, is the general solution to the homogeneous equation:
(D^8 + 6D^6 - 32D^2)y_c = 0
To find the complementary function, we assume a solution of the form:
y_c = e^(rt)
Substituting this into the homogeneous equation, we get:
(r^8 + 6r^6 - 32r^2)e^(rt) = 0
Since e^(rt) is never zero, we can cancel it from both sides of the equation:
r^8 + 6r^6 - 32r^2 = 0
This is a polynomial equation in r, and by solving it, we can find the values of r. The complementary function will then be the linear combination of the solutions e^(rt) with the corresponding values of r.
Remember that solving the polynomial equation is a complex task, and the complementary function will depend on the roots (values of r) obtained from the polynomial equation.
To find the coefficient function of the equation (D^8 + 6D^6 - 32D^2)y = 0, we need to determine the coefficients of each term. Let's break it down:
The term D^8 has coefficient 1 because it is written as D^8.
The term 6D^6 has coefficient 6 because the coefficient in front of D^6 is 6.
The term -32D^2 has coefficient -32 because the coefficient in front of D^2 is -32.
Now, let's write the equation with the coefficients:
1(D^8) + 6(D^6) - 32(D^2)y = 0
Hence, the coefficient function of the equation is 1 + 6 - 32 = -25.
The term D^8 has coefficient 1 because it is written as D^8.
The term 6D^6 has coefficient 6 because the coefficient in front of D^6 is 6.
The term -32D^2 has coefficient -32 because the coefficient in front of D^2 is -32.
Now, let's write the equation with the coefficients:
1(D^8) + 6(D^6) - 32(D^2)y = 0
Hence, the coefficient function of the equation is 1 + 6 - 32 = -25.
To find the complementary function of the given equation, we need to solve for the solution to the homogeneous equation D^8 + 6D^6 - 32D^2 = 0.
First, we can observe that the polynomial equation can be factored as (D^2)(D^6 + 6D^4 - 32) = 0.
Setting D^2 = 0, we find one solution, which is D = 0.
Now, let's focus on the equation D^6 + 6D^4 - 32 = 0. This is a sixth-degree polynomial equation, and it can be difficult to factor or solve directly. However, we can make a change of variable to solve it more easily.
Let u = D^2. The equation then becomes u^3 + 6u^2 - 32 = 0, which is a third-degree polynomial equation.
Using synthetic division or any other method, we find that the solutions to u^3 + 6u^2 - 32 = 0 are u = -4, u = 2, and u = -8.
Now, we can substitute back u = D^2 to find the original values of D:
For u = -4: D^2 = -4, D = ±2i
For u = 2: D^2 = 2, D = ±√2
For u = -8: D^2 = -8, D = ±2√2i
So the solutions to the homogeneous equation D^8 + 6D^6 - 32D^2 = 0 are D = 0, ±2i, ±√2, ±2√2i.
Therefore, the complementary function of the given equation is:
y(x) = c1 + c2e^(2ix) + c3e^(-2ix) + c4e^(√2x) + c5e^(-√2x) + c6e^(2√2ix) + c7e^(-2√2ix) + c8x
First, we can observe that the polynomial equation can be factored as (D^2)(D^6 + 6D^4 - 32) = 0.
Setting D^2 = 0, we find one solution, which is D = 0.
Now, let's focus on the equation D^6 + 6D^4 - 32 = 0. This is a sixth-degree polynomial equation, and it can be difficult to factor or solve directly. However, we can make a change of variable to solve it more easily.
Let u = D^2. The equation then becomes u^3 + 6u^2 - 32 = 0, which is a third-degree polynomial equation.
Using synthetic division or any other method, we find that the solutions to u^3 + 6u^2 - 32 = 0 are u = -4, u = 2, and u = -8.
Now, we can substitute back u = D^2 to find the original values of D:
For u = -4: D^2 = -4, D = ±2i
For u = 2: D^2 = 2, D = ±√2
For u = -8: D^2 = -8, D = ±2√2i
So the solutions to the homogeneous equation D^8 + 6D^6 - 32D^2 = 0 are D = 0, ±2i, ±√2, ±2√2i.
Therefore, the complementary function of the given equation is:
y(x) = c1 + c2e^(2ix) + c3e^(-2ix) + c4e^(√2x) + c5e^(-√2x) + c6e^(2√2ix) + c7e^(-2√2ix) + c8x
From the graph, we can see that there are three real roots: approximately -2.54, -1.46, and 1.31. Therefore, the complementary function of the differential equation is:
y_c = c1 + c2x + c3x^2 + c4x^3 + c5x^4 + c6exp(-2.54x) + c7exp(-1.46x) + c8exp(1.31x)
where c1 to c8 are constants determined by the initial or boundary conditions of the problem.