Find the general solution of x(x^2+6) y'' - 4(x^2+3) y' + 6x…

Find the general solution of $x(x^2+6)\,y'' - 4(x^2+3)\,y' + 6x\,y = 0$.

Step 1: Try $y=x^n$

With $y'=nx^{n-1}$, $y''=n(n-1)x^{n-2}$, substituting gives

\[ n(n-1)x^{n+1}+6n(n-1)x^{n-1} - 4nx^{n+1}-12nx^{n-1} + 6x^{n+1} = 0. \]

The coefficient of the leading power $x^{n+1}$ is $n(n-1)-4n+6=n^2-5n+6=(n-2)(n-3)$, which vanishes for $n=2$ or $n=3$.

Step 2: Check $n=3$

The subdominant power $x^{n-1}$ has coefficient $6n(n-1)-12n=6n(n-3)$, which also vanishes automatically when $n=3$. So $y_1=x^3$ alone already solves the equation exactly. Verify with $y_1'=3x^2$, $y_1''=6x$:

\[ x(x^2+6)(6x) - 4(x^2+3)(3x^2) + 6x\cdot x^3 = 6x^4+36x^2-12x^4-36x^2+6x^4 = 0.\ \checkmark \]

Step 3: Check $n=2$

For $n=2$ the subdominant coefficient is $6(2)(2-3)=-12\neq0$, so $y=x^2$ alone leaves a residual $-12x$; a pure power does not work here. Try a polynomial correction $y=x^2+c$, so $y'=2x$, $y''=2$:

\[ x(x^2+6)(2) - 4(x^2+3)(2x) + 6x(x^2+c) = (2-8+6)x^3+(12-24+6c)x = (6c-12)x. \]

This vanishes for $c=2$, giving the second polynomial solution $y_2=x^2+2$. Verify: $x(x^2+6)(2) - 4(x^2+3)(2x) + 6x(x^2+2) = 2x^3+12x-8x^3-24x+6x^3+12x = 0.$ $\checkmark$

Step 4: Wronskian check

A second-order linear ODE has only a two-dimensional solution space, so $y_1=x^3$ and $y_2=x^2+2$ already form a fundamental system provided their Wronskian is nonzero — no reduction of order is needed here at all:

\[ W(y_1,y_2) = \begin{vmatrix} x^3 & x^2+2 \\ 3x^2 & 2x \end{vmatrix} = 2x^4-3x^2(x^2+2) = -x^4-6x^2 = -x^2(x^2+6) \neq 0 \quad (x\neq0). \]

Step 5: General solution

The solution is recorded in the result below.