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?

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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.