Which of the following are solutions of

2x^2y'' + 5xy' + y = x^2-x on(0,infty)?

Forgot to put the options, didn't know how to go about getting answer, didn't want to just guess.

A. y={x^2}/{15}
B. y= {x^2}/{15}-{x}/{6}
C. y=-{x}/{6}
D. y=x^2-{x}/{6}+{1}/{x}

To find the solutions of the given differential equation, we need to solve the homogeneous equation first, and then find a particular solution.

The homogeneous equation is obtained by setting the right side of the equation equal to zero, i.e., x^2 - x = 0. This equation can be factored as x(x-1) = 0, so the solutions are x = 0 and x = 1.

Next, we can assume a solution of the form y = x^r, where r is a constant. Substituting this into the homogeneous equation, we obtain:

2x^2(r(r-1)) + 5x(r) + x^r = 0

Simplifying the equation, we get:

2r(r-1)x^2 + 5rx + x^r = 0

Since x is not zero, we can divide the equation by x^2 to get:

2r(r-1) + 5r + x^(r-2) = 0

For this equation to hold for all x, each term on the left side must be zero, which gives us two separate equations to solve:

1) 2r(r-1) + 5r = 0
2) x^(r-2) = 0

Solving equation 1) gives us two possible values for r: r = 0 and r = -5/2.

For equation 2), we have x^(r-2) = 0. Since x is not zero, the only way for this equation to hold is if r - 2 = 0, which gives r = 2.

Therefore, the general solution of the homogeneous equation is given by:

y_h = C1x^0 + C2x^(-5/2) + C3x^2

Now, we need to find a particular solution to the non-homogeneous equation 2x^2y'' + 5xy' + y = x^2-x.

To find the particular solution, we can use the method of undetermined coefficients, assuming a particular solution of the form y_p = Ax^3 + Bx^2 + Cx + D, where A, B, C, and D are constants to be determined.

Differentiating y_p twice to find y'' and y', we have:

y''_p = 6Ax + 2B
y'_p = 3Ax^2 + 2Bx + C

Substituting these expressions back into the original non-homogeneous equation, we get:

2x^2(6Ax + 2B) + 5x(3Ax^2 + 2Bx + C) + (Ax^3 + Bx^2 + Cx + D) = x^2 - x

Expanding this equation, we can group the terms with the same powers of x:

(2A)x^3 + (6A + 6B)x^2 + (10A + 10B + C)x + (2B + C + D) = x^2 - x

For this equation to hold for all x, each coefficient of the terms on the left side must be equal to the corresponding coefficient on the right side.

Comparing the coefficients, we have:
2A = 1 (coefficient of x^3)
6A + 6B = 1 (coefficient of x^2)
10A + 10B + C = -1 (coefficient of x)
2B + C + D = 0 (constant term)

Solving this system of equations, we can find the values of A, B, C, and D.

Using the values obtained for A, B, C, and D, the particular solution of the non-homogeneous equation is:

y_p = (1/2)x^3 - (1/2)x^2 - (1/4)x - (1/4)

Finally, the general solution of the non-homogeneous equation is given by:

y = y_h + y_p = C1x^0 + C2x^(-5/2) + C3x^2 + (1/2)x^3 - (1/2)x^2 - (1/4)x - (1/4)

Therefore, the solutions of the given differential equation on (0, ∞) are:

y = C1 + C3x^2 + (1/2)x^3 - (1/2)x^2 - (1/4)x - (1/4)

Note: The values of C1, C2, and C3 can be determined by specifying initial conditions or additional constraints if provided in the problem.