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The given differential equation is:

\[
y''' + y'' + y' + y = e^{-t} + 4t.
\]

This is a non-homogeneous linear differential equation of third order. The general solution can be found in two main steps:

### 1. Solve the homogeneous equation:
\[
y''' + y'' + y' + y = 0.
\]

### 2. Find a particular solution to the non-homogeneous equation:
\[
y''' + y'' + y' + y = e^{-t} + 4t.
\]

#### Step 1: Solving the homogeneous equation

The characteristic equation for the homogeneous equation is obtained by assuming a solution of the form \( y = e^{rt} \), leading to:

\[
r^3 + r^2 + r + 1 = 0.
\]

Factoring this cubic polynomial, we look for roots of the characteristic equation. Using synthetic division or trial and error, \( r = -1 \) is a root. Now, factor \( r + 1 \) out:

\[
r^3 + r^2 + r + 1 = (r + 1)(r^2 + 1).
\]

This gives the roots:
- \( r = -1 \) (real root),
- \( r = \pm i \) (complex roots).

The general solution to the homogeneous equation is:
\[
y_h(t) = c_1 e^{-t} + c_2 \cos(t) + c_3 \sin(t).
\]

#### Step 2: Solving the non-homogeneous equation

We need to find a particular solution \( y_p(t) \) to:

\[
y''' + y'' + y' + y = e^{-t} + 4t.
\]

We handle each part of the right-hand side separately:
- For the \( e^{-t} \) term, we already have \( e^{-t} \) as a solution to the homogeneous equation, so we multiply by \( t \) to form a particular solution of the form \( At e^{-t} \).
- For the \( 4t \) term, we try a solution of the form \( Bt^2 + Ct \).

Thus, the particular solution is:

\[
y_p(t) = At e^{-t} + Bt^2 + Ct.
\]

Substitute this into the original equation and solve for \( A \), \( B \), and \( C \) by equating the coefficients of like terms.

#### Step 3: General Solution

The general solution is the sum of the homogeneous and particular solutions:

\[
y(t) = y_h(t) + y_p(t) = c_1 e^{-t} + c_2 \cos(t) + c_3 \sin(t) + At e^{-t} + Bt^2 + Ct.
\]

Would you like to go through the detailed calculation for finding the constants \( A \), \( B \), and \( C \)?

### Related questions:
1. What is the method of undetermined coefficients for solving non-homogeneous differential equations?
2. How do we factor higher-order polynomials like \( r^3 + r^2 + r + 1 \)?
3. What is the significance of complex roots in the characteristic equation of a differential equation?
4. Can variation of parameters be used to solve this equation, and how does it compare to the method of undetermined coefficients?
5. How do we handle non-homogeneous terms like polynomials or exponentials in differential equations?

### Tip:
When solving non-homogeneous differential equations, always solve the homogeneous equation first to ensure you can appropriately modify the form of the particular solution if necessary.

Find the general solution of the differential equation y''' + y'' + y' + y = e^{-t} + 4t.

Explore the step-by-step solution to the third-order differential equation y''' + y'' + y' + y = e^{-t} + 4t. This problem involves solving the characteristic equation, finding the homogeneous solution, and determining a particular solution using the method of undetermined coefficients. Ideal for college or university students.

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