Find the general solution of y'' + 2y' + y = 3e^(-x)√(x+1)

Find the general solution of $y'' + 2y' + y = 3e^{-x}\sqrt{x+1}$.

Step 1: Homogeneous solution

The characteristic equation is

\[ r^2+2r+1 = (r+1)^2 = 0 \;\Longrightarrow\; r=-1\ (\text{double}), \] so the fundamental system is $\{y_1,y_2\}=\{e^{-x},\,xe^{-x}\}$, and

\[ Y_h = (C_1+C_2x)e^{-x}. \]

Step 2: Variation of parameters — the Wronskian

Seek $y=C_1(x)e^{-x}+C_2(x)xe^{-x}$, with $y_1=e^{-x}$, $y_2=xe^{-x}$. Since $y_2'=e^{-x}-xe^{-x}$, \[W = y_1y_2' - y_1'y_2 = e^{-x}(e^{-x}-xe^{-x}) - (-e^{-x})(xe^{-x}) = e^{-2x} - xe^{-2x} + xe^{-2x} = e^{-2x}.\]

Step 3: Solving for $C_1'(x)$ and $C_2'(x)$

With $g(x)=3e^{-x}\sqrt{x+1}$ (the leading coefficient of $y''$ is already $1$), the variation-of-parameters formulas give

\[ C_1'(x) = -\frac{y_2\,g}{W} = -\frac{xe^{-x}\cdot 3e^{-x}\sqrt{x+1}}{e^{-2x}} = -3x\sqrt{x+1}, \]

\[ C_2'(x) = \frac{y_1\,g}{W} = \frac{e^{-x}\cdot 3e^{-x}\sqrt{x+1}}{e^{-2x}} = 3\sqrt{x+1}. \]

Step 4: Integrating $C_2'(x)$

\[ C_2(x) = 3\int\sqrt{x+1}\,dx = 2(x+1)^{3/2}. \]

Step 5: Integrating $C_1'(x)$

Substitute $u=x+1$ (so $x=u-1$):

\[ \int x\sqrt{x+1}\,dx = \int (u-1)\sqrt{u}\,du = \frac25u^{5/2} - \frac23u^{3/2}. \] So \[C_1(x) = -3\left[\frac25(x+1)^{5/2} - \frac23(x+1)^{3/2}\right] = -\frac65(x+1)^{5/2} + 2(x+1)^{3/2}.\]

Step 6: Assembling the particular solution

With $y_p=C_1(x)e^{-x}+C_2(x)\,xe^{-x}$, factor out $2(x+1)^{3/2}e^{-x}$: $y_p = e^{-x}(x+1)^{3/2}\left[-\frac65(x+1) + 2 + 2x\right].$

Simplify the bracket: \[-\frac65(x+1) + 2 + 2x = \frac45x + \frac45 = \frac45(x+1).\]

Hence $y_p = \frac45 e^{-x}(x+1)^{5/2}.$

Step 7: General solution

The solution is recorded in the result below.

Step 8: Verification

Let $f=(x+1)^{5/2}$, so $f'=\tfrac52(x+1)^{3/2}$ and $f''=\tfrac{15}{4}(x+1)^{1/2}$, and write $y_p=\tfrac45e^{-x}f$. Differentiating a product $e^{-x}f$ twice and substituting into $y_p''+2y_p'+y_p$, the $e^{-x}$ and its derivatives combine as

\[ y_p'' + 2y_p' + y_p = \frac45e^{-x}\Bigl[(f''-2f'+f) + 2(f'-f) + f\Bigr] = \frac45e^{-x}f''. \]

The $f'$ and $f$ terms cancel in this telescoping pattern (characteristic of variation of parameters at a double root), leaving only the $f''$ term: \[\frac45e^{-x}f'' = \frac45e^{-x}\cdot\frac{15}{4}(x+1)^{1/2} = 3e^{-x}\sqrt{x+1}.\ \checkmark\]

The homogeneous terms $(C_1+C_2x)e^{-x}$ satisfy the associated homogeneous equation by construction (they are the fundamental system at the double root $r=-1$), so by linearity the boxed solution satisfies $y''+2y'+y=3e^{-x}\sqrt{x+1}$ for all $C_1,C_2$.