1° Find the general solution of the homogeneous equation…

1° Find the general solution of the homogeneous equation $(H)\!: \qquad L(y) \equiv (2x+1)y'' + (4x-2)y' - 8y = 0$. 2° Find the general solution of the equation with the same left-hand side $(N)\!: \qquad (2x+1)y'' + (4x-2)y' - 8y = (2x+1)\sqrt{x}$.

Step 1: Part 1° — a polynomial solution

Try a quadratic $y_1=ax^2+bx+c$, so $y_1'=2ax+b$, $y_1''=2a$. Substituting into $(H)$ and collecting powers of $x$: $(2x+1)(2a) + (4x-2)(2ax+b) - 8(ax^2+bx+c) = 0$.

Expanding, the $x^2$ terms cancel automatically ($8ax^2-8ax^2=0$), the $x^1$ terms leave $-4b$, and the constant terms leave $2a-2b-8c$. So $b=0$ and, with $b=0$, $c=a/4$. Choosing $a=4$ (hence $c=1$):

\[ y_1 = 4x^2+1. \]

Check: $y_1'=8x$, $y_1''=8$, and $(2x+1)(8)+(4x-2)(8x)-8(4x^2+1)=16x+8+32x^2-16x-32x^2-8=0.\ \checkmark$

Step 2: A second, exponential solution

Try $y_2=e^{-2x}$: $y_2'=-2e^{-2x}$, $y_2''=4e^{-2x}$. Substituting, $(2x+1)(4e^{-2x}) + (4x-2)(-2e^{-2x}) - 8e^{-2x} = e^{-2x}\bigl[8x+4-8x+4-8\bigr] = 0$. $\checkmark$

Step 3: Independence and the general solution of $(H)$

The Wronskian at $x=0$ is $W(0)=y_1(0)y_2'(0)-y_1'(0)y_2(0) = (1)(-2)-(0)(1) = -2 \neq 0$, so $y_1,y_2$ are linearly independent and form a fundamental system.

\[y = C_1(4x^2+1) + C_2e^{-2x}\]

Step 4: Part 2° — standard form and the Wronskian

Dividing $(N)$ by $2x+1$ puts it in standard form $y''+p(x)y'+q(x)y=f(x)$: $y'' + \frac{4x-2}{2x+1}y' - \frac{8}{2x+1}y = \sqrt{x} =: f(x)$.

With $y_1=4x^2+1$ and $y_2=e^{-2x}$ as above, \[W = y_1y_2' - y_1'y_2 = (4x^2+1)(-2e^{-2x}) - 8xe^{-2x} = -2e^{-2x}(4x^2+4x+1) = -2(2x+1)^2e^{-2x}.\]

Step 5: The first quadrature, in closed form

By variation of parameters, $y_p = -y_1\displaystyle\int\frac{y_2f}{W}\,dx + y_2\displaystyle\int\frac{y_1f}{W}\,dx$. For the first term, \[\frac{y_2f}{W} = \frac{e^{-2x}\sqrt{x}}{-2(2x+1)^2e^{-2x}} = -\frac{\sqrt{x}}{2(2x+1)^2}, \qquad\text{so}\qquad -y_1\!\int\!\frac{y_2f}{W}dx = y_1\!\int\!\frac{\sqrt{x}}{2(2x+1)^2}\,dx.\]

Substituting $u=\sqrt{x}$ ($x=u^2$, $dx=2u\,du$) turns this into $\displaystyle\int\frac{u^2}{(2u^2+1)^2}\,du$, a standard rational-in-$u$ integral that evaluates in closed form. Returning to $x$, the result is

\[ I_1(x) := \int\frac{\sqrt{x}}{2(2x+1)^2}\,dx = -\frac{\sqrt{x}}{4(2x+1)} + \frac{\sqrt2}{8}\arctan\!\sqrt{2x}\,, \]

which is confirmed by differentiating $I_1$ directly: $I_1'(x)$ reduces exactly to $\sqrt{x}/[2(2x+1)^2]$. So the first contribution to $y_p$ is

\[ y_1\,I_1(x) = (4x^2+1)\left[-\frac{\sqrt{x}}{4(2x+1)} + \frac{\sqrt2}{8}\arctan\!\sqrt{2x}\right]. \]

Step 6: The second quadrature — a genuine non-elementary term

For the second term, \[\frac{y_1f}{W} = \frac{(4x^2+1)\sqrt{x}}{-2(2x+1)^2e^{-2x}} = -\frac{(4x^2+1)\sqrt{x}\,e^{2x}}{2(2x+1)^2},\]

so that $y_2\displaystyle\int\frac{y_1f}{W}dx = -e^{-2x}I_2(x)$, where

\[ I_2(x) := \int \frac{(4t^2+1)\sqrt{t}\,e^{2t}}{2(2t+1)^2}\,dt. \]

Unlike $I_1$, this integral has no elementary closed form: the combination of a fractional power, a rational function, and an exponential resists every standard technique (substitution, integration by parts, partial fractions), and symbolic computation confirms it cannot be reduced to elementary functions even allowing common special functions. This is not a gap in the calculation — it is a genuine feature of this problem, and the honest way to record $I_2$ is as a quadrature from a fixed basepoint $x_0$ to $x$:

\[ I_2(x) = \int_{x_0}^{x} \frac{(4t^2+1)\sqrt{t}\,e^{2t}}{2(2t+1)^2}\,dt. \]

Step 7: Assembling $y_p$ and the general solution of $(N)$

Combining Steps 5 and 6, and recalling $y_p = -y_1\!\int(y_2f/W)dx + y_2\!\int(y_1f/W)dx = y_1I_1(x) - e^{-2x}I_2(x)$: \[y_p = (4x^2+1)\left[\frac{\sqrt2}{8}\arctan\!\sqrt{2x} - \frac{\sqrt{x}}{4(2x+1)}\right] - e^{-2x}\int_{x_0}^{x}\frac{(4t^2+1)\sqrt{t}\,e^{2t}}{2(2t+1)^2}\,dt.\]

and the general solution of $(N)$ is $y=C_1(4x^2+1)+C_2e^{-2x}+y_p$, i.e.

\[ y = C_1(4x^2+1) + C_2e^{-2x} + (4x^2+1)\left[\frac{\sqrt2}{8}\arctan\!\sqrt{2x} - \frac{\sqrt{x}}{4(2x+1)}\right] - e^{-2x}\!\int_{x_0}^{x}\!\frac{(4t^2+1)\sqrt{t}\,e^{2t}}{2(2t+1)^2}\,dt, \] where $C_1,C_2$ are arbitrary constants and $x_0$ is any fixed basepoint; changing $x_0$ only shifts $I_2$ by a constant, which is absorbed into $C_2$ by linearity, so the solution family does not actually depend on the choice of $x_0$.

\[y = C_1(4x^2+1) + C_2e^{-2x} + (4x^2+1)\!\left[\frac{\sqrt2}{8}\arctan\!\sqrt{2x} - \frac{\sqrt{x}}{4(2x+1)}\right] - e^{-2x}\!\int_{x_0}^{x}\!\frac{(4t^2+1)\sqrt{t}\,e^{2t}}{2(2t+1)^2}\,dt\]