Provide a 3rd order ODE as well as initial conditions at x = 0 such that the
following is the solution to the resulting initial value problem:
y(x) = (2x2 − 2x + 1)e2x

Given the solution \( y(x) = (2x^2 - 2x + 1)e^{2x} \), we need to find a third-order ordinary differential equation (ODE) and corresponding initial conditions at \( x = 0 \).

### Step 1: Compute the derivatives of \( y(x) \)

1. \( y(x) = (2x^2 - 2x + 1)e^{2x} \)

2. First derivative:
   \[
   y'(x) = \frac{d}{dx} \left( (2x^2 - 2x + 1)e^{2x} \right)
   \]
   Use the product rule:
   \[
   y'(x) = (4x - 2)e^{2x} + (2x^2 - 2x + 1)(2e^{2x})
   \]
   Simplify:
   \[
   y'(x) = (4x - 2)e^{2x} + 2(2x^2 - 2x + 1)e^{2x}
   \]
   \[
   y'(x) = \left(4x - 2 + 4x^2 - 4x + 2\right)e^{2x}
   \]
   \[
   y'(x) = \left(4x^2 + 4x - 2\right)e^{2x}
   \]

3. Second derivative:
   \[
   y''(x) = \frac{d}{dx} \left( (4x^2 + 4x - 2)e^{2x} \right)
   \]
   Again, apply the product rule:
   \[
   y''(x) = (8x + 4)e^{2x} + (4x^2 + 4x - 2)(2e^{2x})
   \]
   Simplify:
   \[
   y''(x) = (8x + 4)e^{2x} + 2(4x^2 + 4x - 2)e^{2x}
   \]
   \[
   y''(x) = \left(8x + 4 + 8x^2 + 8x - 4\right)e^{2x}
   \]
   \[
   y''(x) = \left(8x^2 + 16x\right)e^{2x}
   \]

4. Third derivative:
   \[
   y'''(x) = \frac{d}{dx} \left( (8x^2 + 16x)e^{2x} \right)
   \]
   Apply the product rule one more time:
   \[
   y'''(x) = (16x + 16)e^{2x} + (8x^2 + 16x)(2e^{2x})
   \]
   Simplify:
   \[
   y'''(x) = (16x + 16)e^{2x} + 2(8x^2 + 16x)e^{2x}
   \]
   \[
   y'''(x) = \left(16x + 16 + 16x^2 + 32x\right)e^{2x}
   \]
   \[
   y'''(x) = \left(16x^2 + 48x + 16\right)e^{2x}
   \]

### Step 2: Form the third-order ODE

The third-order ODE can be expressed in terms of \( y(x) \), \( y'(x) \), \( y''(x) \), and \( y'''(x) \). Notice that \( y'''(x) \), \( y''(x) \), and \( y'(x) \) all involve powers of \( x \) multiplied by \( e^{2x} \). Comparing the structure, the ODE can be constructed by subtracting the terms to eliminate the powers of \( x \).

We can write the ODE as:
\[
y'''(x) - 6y''(x) + 12y'(x) - 8y(x) = 0
\]
This is the third-order linear homogeneous ODE.

### Step 3: Initial conditions at \( x = 0 \)

Now, let's determine the initial conditions at \( x = 0 \):

- \( y(0) = (2(0)^2 - 2(0) + 1)e^{2(0)} = 1 \)
- \( y'(0) = \left(4(0)^2 + 4(0) - 2\right)e^{2(0)} = -2 \)
- \( y''(0) = \left(8(0)^2 + 16(0)\right)e^{2(0)} = 0 \)

Thus, the initial conditions are:
\[
y(0) = 1, \quad y'(0) = -2, \quad y''(0) = 0
\]

### Final Answer

The third-order ODE is:
\[
y'''(x) - 6y''(x) + 12y'(x) - 8y(x) = 0
\]

The initial conditions at \( x = 0 \) are:
\[
y(0) = 1, \quad y'(0) = -2, \quad y''(0) = 0
\]

Would you like a detailed explanation on any step or a deeper discussion of ODE solutions?

---

Here are 5 related questions:
1. What is the general method to solve a third-order ODE?
2. How does the method of undetermined coefficients apply to solving non-homogeneous ODEs?
3. Can we solve higher-order ODEs using matrix methods?
4. What is the role of initial conditions in determining the uniqueness of a solution?
5. How can variation of parameters be used for non-homogeneous ODEs?

**Tip:** To solve higher-order ODEs, understanding the relationship between characteristic equations and their solutions is key for linear systems.

Provide a 3rd order ODE as well as initial conditions at x = 0 such that the following is the solution to the resulting initial value problem: y(x) = (2x^2 − 2x + 1)e^{2x}

Discover how to construct a third-order ordinary differential equation (ODE) from a given solution y(x) = (2x^2 − 2x + 1)e^{2x}. This guide covers the detailed steps of differentiation, forming the ODE, and determining initial conditions. Ideal for undergraduate students in calculus or differential equations courses.

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