​$$\text{Consider the differential equation:} \\[10pt]x^2 y'' - 2x(x + 1)y' + 2(x + 1)y = 0. \\[10pt]\text{If a polynomial is a solution, then the degree of the polynomial is equal to:}$$​​

Correct option is A

​$$\text{Let } y = a + bx, \text{ then:} \\[10pt]y' = b, \quad y'' = 0. \\[10pt]\text{Substituting into the differential equation:} \\[10pt]x^2 (0) - 2x(x+1)b + 2(x+1)(a + bx) = 0. \\[10pt]\implies -2bx^2 - 2bx + 2ax +2bx^2+ 2a + 2bx = 0. \\[10pt]\implies 2a x + a = 0. \\[10pt]\text{Comparing coefficients, we get:} \\[10pt]a = 0, \quad \text{and } b \text{ is arbitrary.} \\[10pt]\implies y = bx \text{ satisfies the equation and is a solution of degree 1.} \\[10pt]\text{Now, let } y = a + bx + cx^2. \\[10pt]\text{Then, } y' = b + 2cx, \quad y'' = 2c. \\[10pt]\text{Substitute into the differential equation:} \\[10pt]x^2(2c) - 2x(x+1)(b + 2cx) + 2(x+1)(a + bx + cx^2) = 0. \\[10pt]\implies 2cx^2 - 2x(bx+2cx^2 + b+2cx)+2(ax+bx^2+cx^3+a+bx+cx^2)= 0. \\[10pt]\implies x^3(-4c+2c) + x^2(2c-2b-4c+2b+2c) + x(-2b+2b+2a) + 2a = 0. \\[10pt]\text{Comparing coefficients, we get:} \\[10pt]-4c = 0 \implies c = 0, \\[10pt]2a = 0 \implies a = 0, \\[10pt]\text{and } b \text{ is arbitrary.} \\[10pt]\text{Thus, } y = bx \text{ is a solution of degree 1.}\\[10pt] \text{We can check polynomial of degree 3 and 4 also by this method result would be the same.}$$​​