1° Show that (*): xy''' + 4x^2y'' + 4x(2-x^2)y' - 12x^2y…

1° Show that $(*)\!: \qquad xy''' + 4x^2y'' + 4x(2-x^2)y' - 12x^2y - 20x^4 = 0$ is the derivative of a second-order equation. 2° Find the general solution.

Step 1: Finding the antiderivative equation

Try a second-order equation of the form $L[y] \equiv xy''+(4x^2-1)y'-4x^3y = h(x)$ $(**)$, and differentiate its left side with respect to $x$: \[\frac{d}{dx}L[y] = \bigl[y''+xy'''\bigr] + \bigl[8xy'+(4x^2-1)y''\bigr] + \bigl[-12x^2y-4x^3y'\bigr].\]

Collecting terms, the $y''$ contributions combine as $y''+(4x^2-1)y''=4x^2y''$, and the $y'$ contributions as $8xy'-4x^3y'=4x(2-x^2)y'$, so

\[ \frac{d}{dx}L[y] = xy''' + 4x^2y'' + 4x(2-x^2)y' - 12x^2y, \]

which is precisely the left side of $(*)$ apart from the constant term. Matching $(*)$ therefore requires $h'(x)=20x^4$, i.e. $h(x)=4x^5$ (taking the integration constant to be zero; see the closing remark for the effect of a nonzero constant). Hence $(*)$ is exactly the $x$-derivative of

\[ (**)\!: \qquad xy'' + (4x^2-1)y' - 4x^3y = 4x^5. \]

Differentiating the left side of $(**)$ symbolically and comparing with $(*)$ confirms the identity term by term, with $\frac{d}{dx}(4x^5)=20x^4$ matching exactly.\ \checkmark\)

Step 2: A particular solution of $(**)$

Try the polynomial ansatz $y_p=\alpha x^2+\beta$, so $y_p'=2\alpha x$, $y_p''=2\alpha$. Substituting into $(**)$: \[x(2\alpha) + (4x^2-1)(2\alpha x) - 4x^3(\alpha x^2+\beta) = 2\alpha x + \bigl(8\alpha x^3-2\alpha x\bigr) - \bigl(4\alpha x^5+4\beta x^3\bigr).\]

The two $2\alpha x$ terms cancel, leaving

\[ (8\alpha-4\beta)x^3 - 4\alpha x^5 = 4x^5. \]

Matching the $x^5$ coefficients gives $-4\alpha=4$, i.e. $\alpha=-1$; matching the $x^3$ coefficients then gives $8\alpha-4\beta=0$, i.e. $\beta=2\alpha=-2$. Hence

\[ \alpha=-1, \qquad \beta=-2, \qquad y_p = -x^2-2. \]

Direct substitution of $y_p=-x^2-2$ into $xy''+(4x^2-1)y'-4x^3y$ gives exactly $4x^5$, confirming $y_p$.\ \checkmark\)

Step 3: The homogeneous equation and a closed-form reduction

Consider $(H)\!: xy''+(4x^2-1)y'-4x^3y=0$. Substitute $y=e^{-x^2}v(x)$; after simplification the equation for $v$ reduces cleanly, with no first-derivative-times-constant complication, to

\[ (R)\!: \qquad xv'' - v' - 8x^3v = 0. \]

Substituting $y=e^{-x^2}v$ into $(H)$ and dividing through by $e^{-x^2}$ reproduces $(R)$ exactly.\ \checkmark\)

Step 4: Collapsing $(R)$ with $\xi=x^2$

Write $v(x)=V(\xi)$ with $\xi=x^2$, so $v'=2xV'$ and $v''=2V'+4x^2V''=2V'+4\xi V''$. Then

\[ x(2V'+4\xi V'') - 2xV' - 8x^3V = 4x\xi V'' - 8x^3V = 4x\xi V'' - 8x\xi V = 4x\xi\bigl(V''-2V\bigr), \]

using $x^3=x\xi$. Since $4x\xi\neq0$ for $x\neq0$, this forces the simple constant-coefficient equation

\[ V''-2V=0 \;\Longrightarrow\; V(\xi) = Ae^{\sqrt2\,\xi} + Be^{-\sqrt2\,\xi} \;\Longrightarrow\; v(x) = Ae^{\sqrt2\,x^2} + Be^{-\sqrt2\,x^2}. \]

Step 5: The homogeneous solutions

Back-substituting $y=e^{-x^2}v(x)$ gives the two elementary, closed-form homogeneous solutions

\[ y_1 = e^{-x^2}\cdot e^{\sqrt2\,x^2} = e^{(\sqrt2-1)x^2}, \qquad y_2 = e^{-x^2}\cdot e^{-\sqrt2\,x^2} = e^{-(\sqrt2+1)x^2}. \]

Both $y_1$ and $y_2$, substituted directly into the original homogeneous equation $xy''+(4x^2-1)y'-4x^3y$, give a residual of exactly $0$.\ \checkmark\) Their Wronskian, $W = y_1y_2'-y_1'y_2 = -4\sqrt2\,x\,e^{-2x^2}$.

is nonzero for $x\neq0$, confirming $y_1,y_2$ are linearly independent.

Step 6: General solution of $(**)$

Combining the homogeneous pair with the particular solution $y_p=-x^2-2$: $y = C_1e^{(\sqrt2-1)x^2} + C_2e^{-(\sqrt2+1)x^2} - x^2 - 2$.

Substituting this full expression into $xy''+(4x^2-1)y'-4x^3y$ gives exactly $4x^5$ identically in $C_1,C_2$, confirming both the homogeneous pair and the particular solution simultaneously.\ \checkmark\)

\[y = C_1e^{(\sqrt2-1)x^2} + C_2e^{-(\sqrt2+1)x^2} - x^2 - 2\]