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(a). \( \frac{d y}{d x}+2 x y^{2}=0 \) (b). \( \frac{d y}{d x}=e^{3 x+2 y} \)

Ask by Nguyen Rowe. in Colombia
Mar 15,2025

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### (a) Ecuación: \[ \frac{dy}{dx} + 2x\,y^2 = 0 \] **Solución:** \[ y = \frac{1}{x^2 + C} \] --- ### (b) Ecuación: \[ \frac{dy}{dx} = e^{3x+2y} \] **Solución:** \[ e^{-2y} = -\frac{2}{3}\,e^{3x} + C \] donde \( C \) es una constante.

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The Deep Dive

To solve the first differential equation, we identify it as a separable equation: \( \frac{dy}{dx} = -2xy^2 \). We can separate variables by rearranging it to \( \frac{dy}{y^2} = -2x \, dx \). Integrating both sides will yield a solution that involves the logarithm or another algebraic manipulation, potentially leading to a family of curves defined in terms of constants. For the second equation \( \frac{dy}{dx} = e^{3x + 2y} \), this is also separable. Rearranging gives us \( \frac{dy}{e^{2y}} = e^{3x} \, dx \). This allows us to perform an integration on both sides, giving rise to an exponential function in \( y \). The solution will also involve a constant of integration, which can later be used in initial conditions if provided.

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