Question

The value of \( y \) at \( x = 0.3 \) and \( y(x = 0) = 0 \) for the differential equation \[ \frac{dy}{dx} = 3y + 2e^x \] is _________. (Rounded off to 3 decimal places)

Hint:

To solve first-order linear differential equations, use the method of integrating factors, and always apply the initial condition to find the constant of integration.

Answer 1

We are given the first-order linear differential equation: \[ \frac{dy}{dx} = 3y + 2e^x \] with the initial condition \( y(0) = 0 \). We can solve this using the method of integrating factors. First, rewrite the equation: \[ \frac{dy}{dx} - 3y = 2e^x \] The integrating factor is \( e^{\int -3 dx} = e^{-3x} \). Multiply both sides of the differential equation by the integrating factor: \[ e^{-3x} \frac{dy}{dx} - 3e^{-3x} y = 2e^x e^{-3x} \] \[ \frac{d}{dx} \left( e^{-3x} y \right) = 2e^{-2x} \] Integrate both sides with respect to \(x\): \[ e^{-3x} y = \int 2e^{-2x} dx \] \[ e^{-3x} y = -e^{-2x} + C \] Now, multiply through by \( e^{3x} \): \[ y = -e^x + Ce^{3x} \] Using the initial condition \( y(0) = 0 \): \[ 0 = -e^0 + C e^{0} \] \[ 0 = -1 + C \] \[ C = 1 \] Thus, the solution is: \[ y = -e^x + e^{3x} \] Now, substitute \( x = 0.3 \): \[ y(0.3) = -e^{0.3} + e^{0.9} \] \[ y(0.3) \approx -1.3499 + 2.4596 = 1.106 \] Thus, the value of \( y \) at \( x = 0.3 \) is approximately 1.106.