To find the general solution of the differential equation:

\[
y^3 + y'' - y' - y = 0
\]

Let's go through a systematic process.

### Step 1: Consider the Homogeneous Linear Part
The equation contains a mixture of a nonlinear term \( y^3 \) and linear differential terms \( y'', y', \) and \( y \).

The linear part of the equation is:

\[
y'' - y' - y = 0
\]

Let's solve this linear homogeneous equation first.

### Step 2: Solve the Linear Homogeneous Equation
For the linear part \( y'' - y' - y = 0 \), we find the characteristic equation by assuming a solution of the form \( y = e^{rt} \):

\[
r^2 - r - 1 = 0
\]

This is a quadratic equation. Solving for \( r \) using the quadratic formula:

\[
r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}
\]
\[
r = \frac{1 \pm \sqrt{1 + 4}}{2}
\]
\[
r = \frac{1 \pm \sqrt{5}}{2}
\]

So, the roots are:

\[
r_1 = \frac{1 + \sqrt{5}}{2}, \quad r_2 = \frac{1 - \sqrt{5}}{2}
\]

The general solution to the homogeneous linear equation is:

\[
y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}
\]
\[
y_h(t) = C_1 e^{\frac{1 + \sqrt{5}}{2}t} + C_2 e^{\frac{1 - \sqrt{5}}{2}t}
\]

### Step 3: Address the Nonlinear Term \( y^3 \)
Now we need to consider the nonlinear term \( y^3 \) in the original equation:

\[
y^3 + y'' - y' - y = 0
\]

This equation is challenging to solve in general due to the presence of the \( y^3 \) term. Solutions involving the nonlinear term usually require specialized methods such as perturbation techniques, series solutions, or numerical methods.

### Step 4: Consider Simple Solutions or Reductions
As found earlier, the constant solutions \( y = 0 \), \( y = 1 \), and \( y = -1 \) satisfy the equation.

Thus, the general solution might combine these constant solutions with the solution to the homogeneous linear part. However, a fully general solution that addresses the nonlinear term explicitly would require more advanced methods or specific assumptions about the form of the solution.

### Summary of the General Solution
The general solution can be expressed as a combination of the linear homogeneous solution and constant particular solutions:

\[
y(t) = C_1 e^{\frac{1 + \sqrt{5}}{2}t} + C_2 e^{\frac{1 - \sqrt{5}}{2}t}
\]
along with the constant solutions \( y = 0 \), \( y = 1 \), and \( y = -1 \).

If more specific solutions related to the nonlinear part are needed, advanced methods would be required.

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Do you want further details or have any questions?

Here are five related questions that expand on this concept:
1. How can nonlinear terms in differential equations be handled using perturbation methods?
2. What are the applications of characteristic equations in solving differential equations?
3. How can numerical methods be used to approximate solutions to nonlinear differential equations?
4. What are some common nonlinear differential equations and their solutions?
5. How does the superposition principle apply to linear differential equations?

**Tip:** When dealing with mixed linear and nonlinear differential equations, solving the linear part first can provide insights into the structure of the solution.

Learn how to find the general solution of a nonlinear differential equation involving a mixture of nonlinear and linear terms. This solution process includes solving the homogeneous linear part and addressing the challenges posed by the nonlinear term. Suitable for college-level students and those interested in advanced differential equations.

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