91Ó°ÊÓ

The given initial value problem (IVP) consists of a first-order ordinary differential equation (ODE) and an initial condition: ODE: \(y'(x) = -y + 2\) Initial condition: \(y(0) = -2\)

The first-order ODE given is a linear ODE, specifically \(y'(x) + y(x) = 2\). To solve this ODE, first, we calculate the integrating factor, which is given by \(e^{\int px dx}\), where \(px\) is the coefficient of \(y\). In this case, \(px = 1\). So, the integrating factor is: \(e^{\int 1 dx} = e^x\). Now, multiply both sides of the ODE by the integrating factor: \(e^x y'(x) + e^x y(x) = 2 e^x\) The left-hand side of the above equation is the derivative of the product of \(e^x\) and \(y(x)\): \(\dfrac{d}{dx} (e^x y(x)) = 2 e^x\) Now integrate both sides with respect to \(x\): \(\int \dfrac{d}{dx} (e^x y(x)) dx = \int 2 e^x dx\) This results in: \(e^x y(x) = 2 e^x + C\) Now, divide both sides by \(e^x\) to find the general solution of the ODE: \(y(x) = 2 + \dfrac{C}{e^x}\)

Substitute the initial condition \(y(0) = -2\) into the general solution to find the constant \(C\): \(-2 = 2 + \dfrac{C}{e^0}\) Solve for \(C\): \(-4 = C\)

Substitute the value of \(C\) back into the general solution to obtain the particular solution of the IVP: \(y(x) = 2 - \dfrac{4}{e^x}\) The particular solution for the given initial value problem is \(y(x) = 2 - \dfrac{4}{e^x}\).