solve the differential equation by variation of parameters. $y + 3y + 2y=\frac{1}{3 + e^{x}}$

Explanation:

Step1: Find the complementary - function

The homogeneous equation is $y'' + 3y'+2y = 0$. The characteristic equation is $r^{2}+3r + 2=0$. Factoring gives $(r + 1)(r+2)=0$. So, $r_1=-1$ and $r_2=-2$. The complementary - function $y_c = C_1e^{-x}+C_2e^{-2x}$.

Step2: Assume the particular - solution form

We assume a particular solution of the form $y_p=u_1(x)e^{-x}+u_2(x)e^{-2x}$. We have $W=\begin{vmatrix}e^{-x}&e^{-2x}\-e^{-x}&-2e^{-2x}\end{vmatrix}=-2e^{-3x}+e^{-3x}=-e^{-3x}$. $W_1=\begin{vmatrix}0&e^{-2x}\\frac{1}{3 + e^{x}}&-2e^{-2x}\end{vmatrix}=-\frac{e^{-2x}}{3 + e^{x}}$ and $W_2=\begin{vmatrix}e^{-x}&0\-e^{-x}&\frac{1}{3 + e^{x}}\end{vmatrix}=\frac{e^{-x}}{3 + e^{x}}$.

Step3: Calculate $u_1(x)$ and $u_2(x)$

$u_1(x)=\int\frac{W_1}{W}dx=\int\frac{e^{-2x}}{(3 + e^{x})e^{-3x}}dx=\int\frac{e^{x}}{3 + e^{x}}dx=\ln(3 + e^{x})+C_3$. $u_2(x)=\int\frac{W_2}{W}dx=\int\frac{-e^{-x}}{(3 + e^{x})e^{-3x}}dx=-\int\frac{e^{2x}}{3 + e^{x}}dx$. Let $t = e^{x}$, then $dt=e^{x}dx$ and the integral becomes $-\int\frac{t}{3 + t}dt=-\int(1-\frac{3}{3 + t})dt=-t + 3\ln(3 + t)+C_4=-e^{x}+3\ln(3 + e^{x})+C_4$.

Step4: Find the particular - solution

$y_p=\ln(3 + e^{x})e^{-x}+(-e^{x}+3\ln(3 + e^{x}))e^{-2x}$.

Step5: Find the general solution

The general solution $y(x)=y_c + y_p=C_1e^{-x}+C_2e^{-2x}+\ln(3 + e^{x})e^{-x}+(-e^{x}+3\ln(3 + e^{x}))e^{-2x}$.

Answer:

$y(x)=C_1e^{-x}+C_2e^{-2x}+\ln(3 + e^{x})e^{-x}+(-e^{x}+3\ln(3 + e^{x}))e^{-2x}$