Consider (*): y'' + (4x-2)/(2x+1)y' - (8)/(2x+1)y =…

Consider $(*)\!: \qquad y'' + \frac{4x-2}{2x+1}y' - \frac{8}{2x+1}y = 2(2x+1)^2e^{-x}$, whose homogeneous part has a particular solution of the form $y=e^{mx}$. 1° Find the general solution of $(*)$. 2° Find the integral curve tangent to the $x$-axis at the origin, i.e. satisfying $y(0)=0$, $y'(0)=0$.

Step 1: Part 1° — the homogeneous solutions

Multiplying the homogeneous part of $(*)$ by $2x+1$ gives $(2x+1)y''+(4x-2)y'-8y=0$, exactly the operator $L(y)$ of the preceding problem. Its fundamental system is again

\[ y_1 = 4x^2+1, \qquad y_2 = e^{-2x}, \]

so the exponential solution promised in the statement is $y_2=e^{mx}$ with $m=-2$, and the Wronskian is again

\[ W = y_1y_2' - y_1'y_2 = -2(2x+1)^2e^{-2x}. \]

Step 2: Variation of parameters

Equation $(*)$ is already in standard form, with right-hand side $f(x)=2(2x+1)^2e^{-x}$. Compute the two integrands: \[\frac{y_2f}{W} = \frac{e^{-2x}\cdot2(2x+1)^2e^{-x}}{-2(2x+1)^2e^{-2x}} = -e^{-x}, \qquad \frac{y_1f}{W} = \frac{(4x^2+1)\cdot2(2x+1)^2e^{-x}}{-2(2x+1)^2e^{-2x}} = -(4x^2+1)e^{x}.\]

Integrating each (dropping the arbitrary constants, which only reproduce multiples of $y_1,y_2$): \[A(x) := \int\frac{y_2f}{W}\,dx = e^{-x}, \qquad B(x) := \int\frac{y_1f}{W}\,dx = \int-(4x^2+1)e^x\,dx = (-4x^2+8x-9)e^x,\]

the second by two integrations by parts on $\int x^2e^x\,dx$ and $\int e^x\,dx$.

Step 3: Assembling $y_p$ and the general solution

With $y_p = -y_1A + y_2B$: $y_p = -(4x^2+1)e^{-x} + e^{-2x}(-4x^2+8x-9)e^{x} = -(4x^2+1)e^{-x} + (-4x^2+8x-9)e^{-x}$.

Combining the two $e^{-x}$-terms, $y_p = (-8x^2+8x-10)e^{-x} = -2(4x^2-4x+5)e^{-x}$.

so the general solution of $(*)$ is

\[ y = C_1(4x^2+1) + C_2e^{-2x} - 2(4x^2-4x+5)e^{-x}. \]

Step 4: Verification of $y_p$

Substituting $y_p=-2(4x^2-4x+5)e^{-x}$ into the standard-form left-hand side of $(*)$ reproduces $2(2x+1)^2e^{-x}$ exactly, and the full general solution satisfies $(*)$ identically in $C_1,C_2$, confirming both the particular solution and that $y_1,y_2$ are unaffected (they solve the homogeneous part).

\[y = C_1(4x^2+1) + C_2e^{-2x} - 2(4x^2-4x+5)e^{-x}\]

Step 5: Part 2° — the curve tangent to the $x$-axis at the origin

Differentiating the general solution (writing $g(x)=4x^2-4x+5$, $g'(x)=8x-4$, so that the derivative of $-2ge^{-x}$ is $2(g-g')e^{-x}=2e^{-x}(4x^2-12x+9)$): $y' = 8C_1x - 2C_2e^{-2x} + 2e^{-x}\bigl(4x^2-12x+9\bigr)$.

At $x=0$: $y_p(0)=-2(5)=-10$, so $y(0)=C_1+C_2-10=0$. Also $y'(0)=-2C_2+2(9)=-2C_2+18=0$, giving $C_2=9$, and then $C_1=10-C_2=1$.

Step 6: Verification

With $C_1=1,\,C_2=9$: $y(0)=1+9-10=0\ \checkmark$, and $y'(0)=8(1)(0)-2(9)+2(9)=0-18+18=0\ \checkmark$.

\[y = (4x^2+1) + 9e^{-2x} - 2(4x^2-4x+5)e^{-x}\]