Question

After Verifying that the functions 1 2 satisfy the corresponding homogeneous equation of the given equation, find a particular solution of the non-homogeneous equation and then the general solution of the equation .

x²y'' + xy' + (x² - 0.25 ) y = 3x √xsinx



x> 0
y1(x) = sin (x) / √x
y2(x) = cos (x) / √x​

Answer 1

To find a particular solution of the non-homogeneous equation and the general solution of the equation, we can use the method of variation of parameters.

First, let's find the Wronskian of the homogeneous solutions y1(x) and y2(x):

W(y1, y2) = | y1 y2 |

| y1' y2' |

We have y1(x) = sin(x) / √x and y2(x) = cos(x) / √x. Differentiating these functions, we get:

y1'(x) = (cos(x) / √x - sin(x) / (2√x^3))

y2'(x) = (-sin(x) / √x - cos(x) / (2√x^3))

Substituting these values into the Wronskian:

W(y1, y2) = | sin(x) / √x cos(x) / √x |

| (cos(x) / √x - sin(x) / (2√x^3)) (-sin(x) / √x - cos(x) / (2√x^3)) |

Expanding the determinant:

W(y1, y2) = (sin(x) / √x) * (-sin(x) / √x - cos(x) / (2√x^3)) - (cos(x) / √x) * (cos(x) / √x - sin(x) / (2√x^3))

Simplifying:

W(y1, y2) = -1 / (2√x)

Now, we can find the particular solution using the variation of parameters formula:

y_p(x) = -y1(x) * ∫(y2(x) * g(x)) / W(y1, y2) dx + y2(x) * ∫(y1(x) * g(x)) / W(y1, y2) dx

Here, g(x) = 3x√xsin(x). Substituting the values:

y_p(x) = -((sin(x) / √x) * ∫((3x√xsin(x)) * (-1 / (2√x))) dx + (cos(x) / √x) * ∫((3x√xsin(x)) / (2√x)) dx

Simplifying the integrals:

y_p(x) = -(∫(-3sin^2(x)) dx) + (∫(3xcos(x)sin(x)) dx)

Integrating:

y_p(x) = 3/2 (xsin^2(x) - cos^2(x)) - 3/2 (xcos^2(x) + sin^2(x)) + C

Simplifying:

y_p(x) = 3x(sin^2(x) - cos^2(x)) + C

The general solution of the equation is given by the sum of the homogeneous solutions and the particular solution:

y(x) = C1 * (sin(x) / √x) + C2 * (cos(x) / √x) + 3x(sin^2(x) - cos^2(x)) + C

where C1, C2, and C are arbitrary constants.