Question

Solve the initial-value problem x(2 + x)y' + 2(1 + x)y = 1 + 3x^2, y(-1) = 1.

Answer 1

To solve the given initial-value problem, use the method of integrating factors and solve the resulting differential equation. Don't forget to apply the initial condition to find the specific solution.

Step-by-step explanation:

To solve the initial-value problem x(2 + x)y' + 2(1 + x)y = 1 + 3x^2, y(-1) = 1, we can use the method of integrating factors.

1. First, rearrange the equation to get it in standard form: y' + (2/(2 + x))(1 + x)y = (1 + 3x^2)/(2 + x).
2. Identify the integrating factor as e^(∫(2/(2 + x))dx). Solve the integral to get the integrating factor as e^(2ln|2 + x|) = (2 + x)^2.
3. Multiply the entire equation by the integrating factor (2 + x)^2.
4. Now, the equation becomes (2 + x)^2y' + 2(1 + x)(2 + x)^2y = (1 + 3x^2)(2 + x)^2.
5. This equation can be simplified further and solved using standard methods, such as separation of variables.
6. Finally, plug in the initial condition y(-1) = 1 to find the specific solution.