Use any two methods of separation of variables, 1st order linear,homogenous, Bernoulli, exact, etc to find the GS of𝑦 " = 2𝑥𝑦 + 2𝑥𝑥 # + 1

To solve the second-order differential equation $y'' = 2xy + 2x^2 + 1$, we'll employ two different methods for solving differential equations. First, let's rewrite the equation in a more manageable form.

1. Break Down the Problem

The given equation is:

$y'' - 2xy = 2x^2 + 1$

This is a non-homogeneous linear differential equation. We will first find the complementary solution by solving the associated homogeneous equation and then find a particular solution.

2. Relevant Concepts

Step 1: Complementary Solution (Homogeneous Equation)

The homogeneous part of the equation is:

$y'' - 2xy = 0$

This is a second-order linear homogeneous differential equation.

Step 2: Particular Solution

For the non-homogeneous part $2x^2 + 1$, we will use the method of undetermined coefficients to find a particular solution.

3. Analysis and Detail

Step 1: Solve the Homogeneous Equation

1. The characteristic equation for the homogeneous part is obtained by assuming solutions of the form $y = e^{rx}$ or through series solutions.
2. For this equation, we can try a solution in the form of power series or special functions like Hermite polynomials due to the nature of the coefficients.

Assuming a solution of the form $y = c_1 e^{x^2} + c_2 e^{-x^2}$.

However, a more straightforward method is to recognize the structure and solve it directly with special functions or polynomial expansions.

Step 2: Finding the Particular Solution

We will guess a solution of the form:

$y_p = Ax^2 + Bx + C$

Now, we need to compute the first and second derivative:

$y_p' = 2Ax + B$ $y_p'' = 2A$

Substituting $y_p$ and its derivatives into the original equation:

$2A - 2x(Ax^2 + Bx + C) = 2x^2 + 1$ This gives: $2A - 2Ax^2 - 2Bx - 2C = 2x^2 + 1$

Equating coefficients:

1. For $x^2: -2A = 2$ $\implies A = -1$
2. For $x: -2B = 0$ $\implies B = 0$
3. For the constant term: $2A - 2C = 1$ $\implies -2 - 2C = 1$ $\implies C = -\frac{3}{2}$

Thus, the particular solution is:

$y_p = -x^2 - \frac{3}{2}$

4. Verify and Summarize

Now, the general solution combines the homogeneous and particular solutions:

$y = C_1 e^{x^2} + C_2 e^{-x^2} - x^2 - \frac{3}{2}$

Final Answer

The general solution of the differential equation $y'' = 2xy + 2x^2 + 1$ is:

$y = C_1 e^{x^2} + C_2 e^{-x^2} - x^2 - \frac{3}{2}$

where $C_1$ and $C_2$ are arbitrary constants.