Question

Solve

x ^ 2 * (d ^ 2 * y)/(d * x ^ 2) - 2x * (dy)/(dx) - 4y = x ^ 2 + 2 * log x

Answer 1

$$x^2y''-2xy'-4y=x^2+2\ln x$$

The corresponding homogeneous differental equation

$$x^2y''-2xy'-4y=0$$

Auxiliary equation

$$x^2(x^r)''-2x(x^r)'-4x^r=0$$

$$x^r(r(r-1)-2r-4)=0$$

$$r^2-3r-4=0$$

$$(r+1)(r-4)=0$$

$$r_1=-1, r_2=4$$

The general solution of the homogeneous differential equation is

$$y_h=\dfrac{c_1}{x}+c_2x^4$$

Find the particular solution of the non-homogeneous differential equation

$$y_p=Ax^2+Bx+C+D\ln x$$

$$y_p'=2Ax+B+\dfrac{D}{x}$$

$$y_p''=2A-\dfrac{D}{x^2}$$

Substitute

$$2Ax^2-D-4Ax^2-2Bx-2D$$

$$-4Ax^2-4Bx-4C-4D\ln x=x^2+2\ln x$$

$$-6A=1$$

$$-6B=0$$

$$-3D-4C=0$$

$$D=-\dfrac{1}{2}$$

$$A=-\dfrac{1}{6}, B=0, C=\dfrac{3}{8}$$

$$y_p=-\dfrac{1}{6}x^2+\dfrac{3}{8}-\dfrac{1}{2}\ln x$$

The general solution of the given non-homogeneous differential equation is

$$y=\dfrac{c_1}{x}+c_2x^4-\dfrac{1}{6}x^2+\dfrac{3}{8}-\dfrac{1}{2}\ln x$$