Answer: It is stated down below
Explanation:
To solve the given second-order linear homogeneous differential equation, we can use the method of undetermined coefficients. Let's solve it step by step:
The given differential equation is:
x^2y'' + 6xy' - 14y = x^3
We assume a particular solution of the form y_p(x) = Ax^3, where A is a constant to be determined.
Now, let's find the first and second derivatives of y_p(x):
y_p'(x) = 3Ax^2
y_p''(x) = 6Ax
Substituting these derivatives back into the differential equation:
x^2(6Ax) + 6x(3Ax^2) - 14(Ax^3) = x^3
Simplifying the equation:
6Ax^3 + 18Ax^3 - 14Ax^3 = x^3
10Ax^3 = x^3
Now, comparing the coefficients on both sides of the equation:
10A = 1
A = 1/10
So, the particular solution is y_p(x) = (1/10)x^3.
To find the general solution, we need to consider the complementary solution to the homogeneous equation, which satisfies the equation:
x^2y'' + 6xy' - 14y = 0
We can solve this homogeneous equation by assuming a solution of the form y_c(x) = x^r, where r is a constant to be determined.
Differentiating y_c(x) twice:
y_c'(x) = rx^(r-1)
y_c''(x) = r(r-1)x^(r-2)
Substituting these derivatives back into the homogeneous equation:
x^2(r(r-1)x^(r-2)) + 6x(rx^(r-1)) - 14x^r = 0
Simplifying the equation:
r(r-1)x^r + 6rx^r - 14x^r = 0
(r^2 - r + 6r - 14)x^r = 0
(r^2 + 5r - 14)x^r = 0
For this equation to hold for all values of x, the coefficient (r^2 + 5r - 14) must be equal to zero. So we solve:
r^2 + 5r - 14 = 0
Factoring the equation:
(r + 7)(r - 2) = 0
This gives two possible values for r:
r_1 = -7
r_2 = 2
Therefore, the complementary solution is y_c(x) = C_1x^(-7) + C_2x^2, where C_1 and C_2 are constants.
The general solution is given by the sum of the particular and complementary solutions:
y(x) = y_p(x) + y_c(x)
= (1/10)x^3 + C_1x^(-7) + C_2x^2
To find the values of C_1 and C_2, we can use the initial conditions:
y(1) = 3
y'(1) = 3
Substituting these values into the general solution:
3 = (1/10)(1)^3 + C_1(1)^(-7) + C_2(1)^2
3 = 1/10 + C_1 + C_2
3 = 1/10 + C_1 + C_2 (Equation 1)
3 = (3/10) + C_1 + 1(C_2) (Equation 2)
From Equation 1, we get:
C_1 + C_2 = 3 - 1/10
From Equation 2, we get:
C_1 + C_2 = 3 - 3/10
Combining the equations:
C_1 + C_2 = 27/10 - 3/10
C_1 + C_2 = 24/10
C_1 + C_2 = 12/5
Since C_1 + C_2 is a constant, we can represent it as another constant, let's call it C.
C_1 + C_2 = C
Therefore, the general solution can be written as:
y(x) = (1/10)x^3 + C_1x^(-7) + C_2x^2
= (1/10)x^3 + Cx^(-7) + Cx^2
Thus, y as a function of x is given by:
y(x) = (1/10)x^3 + Cx^(-7) + Cx^2, where C is a constant.