Questions: Differential Equations - II Solve the differential equation with the given initial condition. dy/dx=-2 y (0,3) y=[?] e^[ ]

Solution Steps

To solve the differential equation \( \frac{dy}{dx} = -2y \), we first rewrite it in the standard form: \[ \frac{dy}{dx} + 2y = 0 \] The integrating factor \( \mu(x) \) is calculated as follows: \[ \mu(x) = e^{\int 2 \, dx} = e^{2x} \]

We multiply the entire differential equation by the integrating factor: \[ e^{2x} \frac{dy}{dx} + 2y e^{2x} = 0 \] This can be rewritten as: \[ \frac{d}{dx} \left( y e^{2x} \right) = 0 \]

Integrating both sides gives: \[ \int \frac{d}{dx} \left( y e^{2x} \right) dx = \int 0 \, dx \] This results in: \[ y e^{2x} = C \] where \( C \) is the constant of integration.

Rearranging the equation, we find the general solution: \[ y(x) = C e^{-2x} \]

Using the initial condition \( y(0) = 3 \): \[ 3 = C e^{-2 \cdot 0} \implies 3 = C \] Thus, \( C = 3 \).

Substituting \( C \) back into the general solution, we obtain the particular solution: \[ y(x) = 3 e^{-2x} \]

Final Answer