The solution of y' - y = 1, y(0) = -1 is give | vedclass

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -1$ and $Q(x) = 1$.The

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -1$ and $Q(x) = 1$.The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$.Multiplying both sides by the $IF$,we get $e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x}$,which simplifies to $\frac{d}{dx}(y e^{-x}) = e^{-x}$.Integrating both sides with respect to $x$,we get $y e^{-x} = \int e^{-x} dx = -e^{-x} + C$.Thus,$y = -1 + C e^x$.Using the initial condition $y(0) = -1$,we substitute $x = 0$ and $y = -1$:$-1 = -1 + C e^0 \Rightarrow -1 = -1 + C \Rightarrow C = 0$.Substituting $C = 0$ back into the general solution,we get $y(x) = -1$.

The solution of $y' - y = 1, y(0) = -1$ is given by $y(x) = $

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