Solve the differential equation: \[ x^2y \, dx - (x^3 + y^3) \, dy = 0. \]

Question

Given that the solution of \[ \frac{d^2y}{dx^2} + \alpha \frac{dy}{dx} + \beta y = -e^{-x} \] is \[ y(x) = C_1 e^{-x} + C_2 e^{2x} + x e^{-x}, \] find the values of $\alpha$ and $\beta$.

Answer 1

The differential equation $ x^2y \, dx - (x^3 + y^3) \, dy = 0 $ is provided. Rearranging, we get $ x^2y \, dx = (x^3 + y^3) \, dy $. Dividing both sides by $x^2y(x^3 + y^3)$ yields $ \frac{dx}{x^3 + y^3} = \frac{dy}{x^2y} $. This equation is separable. Integrating both sides gives $ \int \frac{dx}{x^3 + y^3} = \int \frac{dy}{x^2y} $. The evaluation of the left-hand integral may necessitate advanced techniques such as substitution or numerical methods. The general solution can be stated as $ F(x, y) = C $, where $F(x, y)$ represents the potential function obtained from integration, and $C$ is the integration constant.

Solve the differential equation: \[ x^2y \, dx - (x^3 + y^3) \, dy = 0. \]

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Solution and Explanation

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