We are given the differential equation $y'' + 3y' + 2y = \frac{x-1}{x^2}e^{-x}$ (E). 1. We need to show that $p(x) = e^{-x} \ln x$ is a particular solution of (E).

1. Problem Description

We are given the differential equation $y'' + 3y' + 2y = \frac{x-1}{x^2}e^{-x}$ (E).

1. We need to show that $p(x) = e^{-x} \ln x$ is a particular solution of (E).

2. We need to prove that a function $f$, defined on $]0, +\infty[$, is a solution of the differential equation (E) if and only if $f - p$ is a solution of the differential equation $y'' + 3y' + 2y = 0$ (E').

3. We need to find the solutions of the differential equation $y'' + 3y' + 2y = 0$ (E').

4. We need to find all solutions of the differential equation (E).

2. Solution Steps

1. We need to show that $p(x) = e^{-x} \ln x$ is a solution of $y'' + 3y' + 2y = \frac{x-1}{x^2}e^{-x}$.

First, compute the first derivative of $p(x)$:

$p'(x) = -e^{-x} \ln x + e^{-x} \frac{1}{x} = e^{-x}(\frac{1}{x} - \ln x)$.

Next, compute the second derivative of $p(x)$:

$p''(x) = -e^{-x}(\frac{1}{x} - \ln x) + e^{-x}(-\frac{1}{x^2} - \frac{1}{x}) = e^{-x}(-\frac{1}{x} + \ln x - \frac{1}{x^2} - \frac{1}{x}) = e^{-x}(\ln x - \frac{2}{x} - \frac{1}{x^2})$.

Now, we substitute $p$, $p'$, and $p''$ into the differential equation:

$p'' + 3p' + 2p = e^{-x}(\ln x - \frac{2}{x} - \frac{1}{x^2}) + 3e^{-x}(\frac{1}{x} - \ln x) + 2e^{-x} \ln x = e^{-x}(\ln x - \frac{2}{x} - \frac{1}{x^2} + \frac{3}{x} - 3\ln x + 2\ln x) = e^{-x}(\frac{1}{x} - \frac{1}{x^2}) = e^{-x}(\frac{x - 1}{x^2})$.

Thus, $p'' + 3p' + 2p = \frac{x-1}{x^2} e^{-x}$, which means that $p(x) = e^{-x} \ln x$ is indeed a solution of the differential equation (E).

2. Let $f$ be a solution of (E), i.e., $f'' + 3f' + 2f = \frac{x-1}{x^2}e^{-x}$. If $f - p$ is a solution to $y'' + 3y' + 2y = 0$ (E'), then $(f-p)'' + 3(f-p)' + 2(f-p) = 0$. This gives us $f'' - p'' + 3f' - 3p' + 2f - 2p = 0$, so $f'' + 3f' + 2f = p'' + 3p' + 2p$. But we know that $p'' + 3p' + 2p = \frac{x-1}{x^2} e^{-x}$, so $f'' + 3f' + 2f = \frac{x-1}{x^2} e^{-x}$, which means that $f$ is a solution of (E).

Conversely, if $f$ is a solution to (E), i.e., $f'' + 3f' + 2f = \frac{x-1}{x^2}e^{-x}$, then $(f-p)'' + 3(f-p)' + 2(f-p) = f'' - p'' + 3(f' - p') + 2(f - p) = f'' + 3f' + 2f - (p'' + 3p' + 2p) = \frac{x-1}{x^2}e^{-x} - \frac{x-1}{x^2}e^{-x} = 0$. Thus, $f - p$ is a solution of $y'' + 3y' + 2y = 0$.

3. We need to solve $y'' + 3y' + 2y = 0$. The characteristic equation is $r^2 + 3r + 2 = 0$, which factors as $(r+1)(r+2) = 0$. Thus, the roots are $r = -1$ and $r = -2$. Therefore, the solutions are of the form $y(x) = c_1 e^{-x} + c_2 e^{-2x}$, where $c_1$ and $c_2$ are constants.

4. Since $f - p$ is a solution of $y'' + 3y' + 2y = 0$, we have $f(x) - p(x) = c_1 e^{-x} + c_2 e^{-2x}$. Thus, $f(x) = p(x) + c_1 e^{-x} + c_2 e^{-2x}$. Substituting $p(x) = e^{-x} \ln x$, we get $f(x) = e^{-x} \ln x + c_1 e^{-x} + c_2 e^{-2x}$, where $c_1$ and $c_2$ are constants.

3. Final Answer

1. $p(x) = e^{-x} \ln x$ is a solution of equation (E).

2. $f$ is a solution of (E) if and only if $f - p$ is a solution of (E').

3. The solutions of (E') are $y(x) = c_1 e^{-x} + c_2 e^{-2x}$.

4. The solutions of (E) are $f(x) = e^{-x} \ln x + c_1 e^{-x} + c_2 e^{-2x}$, where $c_1$ and $c_2$ are constants.