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resuelve por metodo de coeficientes indeterminados

The differential equation given is:

\[
y'' + 2y' - 3y = 1 + xe^x
\]

We will solve this using the method of undetermined coefficients. This method is applied to non-homogeneous linear differential equations, where the right-hand side (non-homogeneous part) is a combination of functions like polynomials, exponentials, and trigonometric functions.

### Steps to solve:

#### 1. Solve the homogeneous equation:
First, solve the homogeneous equation associated with the given differential equation:

\[
y_h'' + 2y_h' - 3y_h = 0
\]

This is a second-order linear homogeneous equation with constant coefficients. The characteristic equation is:

\[
r^2 + 2r - 3 = 0
\]

Solve for \(r\) using the quadratic formula:

\[
r = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2}
\]

So the roots are:

\[
r_1 = 1 \quad \text{and} \quad r_2 = -3
\]

Thus, the general solution to the homogeneous equation is:

\[
y_h(x) = c_1e^x + c_2e^{-3x}
\]

#### 2. Find the particular solution \(y_p\):
For the non-homogeneous part \(1 + xe^x\), we propose a particular solution based on the form of the right-hand side. 

- For the constant "1", we try a constant solution: \(A\).
- For the term \(xe^x\), we try the form \(x(Bx + C)e^x\), since \(e^x\) already appears in the homogeneous solution.

Thus, the proposed particular solution is:

\[
y_p(x) = A + (Bx^2 + Cx)e^x
\]

#### 3. Differentiate \(y_p\):
First, calculate \(y_p'\) and \(y_p''\).

\[
y_p' = (2Bx + C)e^x + (Bx^2 + Cx)e^x
= (2Bx + C + Bx^2 + Cx)e^x
\]

\[
y_p'' = (2B + 2Bx + C)e^x + (2Bx + C + Bx^2 + Cx)e^x
= (2B + 4Bx + 2Cx + Bx^2)e^x
\]

#### 4. Substitute into the original equation:
Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original equation \(y'' + 2y' - 3y = 1 + xe^x\):

\[
(2B + 4Bx + 2Cx + Bx^2)e^x + 2((2Bx + C + Bx^2 + Cx)e^x) - 3(A + (Bx^2 + Cx)e^x) = 1 + xe^x
\]

Group terms with respect to powers of \(x\) and \(e^x\):

\[
(A + terms \, in \, x)e^x = 1 + xe^x
\]

Equating coefficients of \(x^2e^x\), \(xe^x\), and \(e^x\) gives a system of equations to solve for \(A\), \(B\), and \(C\).

Let me now compute the values of \(A\), \(B\), and \(C\).It seems that the equation-solving process returned some terms with fractions that need to be properly balanced. Let's break it down manually to match coefficients of powers of \(x\) and \(e^x\).

Upon comparison of terms from the expanded equation with the right-hand side \(1 + xe^x\), we can equate the coefficients for powers of \(x^2\), \(x\), and constants to solve for \(A\), \(B\), and \(C\). Let me refine this computation manually to clarify the final coefficients.It seems like I can't do more advanced data analysis right now. Please try again later. However, you can continue solving this by equating the coefficients of \(e^x\) from the expanded equation to find the values of \(A\), \(B\), and \(C\). Let me know if you'd like help with further steps or clarifications!

Here are some follow-up questions that can help deepen understanding:

1. How is the method of undetermined coefficients different from variation of parameters?
2. What is the significance of the homogeneous solution in the overall solution?
3. Can the method of undetermined coefficients handle non-polynomial terms, and if so, how?
4. How do we decide the form of the particular solution for more complex right-hand sides?
5. What are the possible adjustments when the right-hand side involves terms already in the homogeneous solution?

**Tip**: Always verify if any term from your particular solution overlaps with the homogeneous one; this requires multiplication by \(x\) if needed.

Solve the differential equation y'' + 2y' - 3y = 1 + xe^x using the method of undetermined coefficients.

Learn how to solve the differential equation y'' + 2y' - 3y = 1 + xe^x using the method of undetermined coefficients. This step-by-step guide covers key concepts in linear differential equations, characteristic equations, and finding particular solutions.

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