Solving IVPs with one-sided initial condition using Laplace Transform

Question

Consider an initial value problem of first order linear ODE. $$\frac{\mathrm{d}}{\mathrm{d}t}y(t)+2y(t)=e^{-t}H(t),\lim_{t \to 0^-}y(t)=2$$ where $H(t)$ is the Heaviside function. Let $L[y: t \to s]=Y(s)$ and applying Laplace Transform to both side of the equation. $$sY(s)-y(0)+2Y(s)=\frac{1}{s+1}.$$ Then I find $y(t)$ to be $$Y(s)=\frac{1}{(s+1)(s+2)}+\frac{1}{s+2}y(0) \Rightarrow y(t)=y(0)e^{-2t}+(e^{-t}-e^{-2t}).$$ My main question is whether it is appropriate to use $y(0^-)$ in this case as the value of $y(0)$, which in this case only the value of left sided limit is given.

Answer 1

Note that the solution $y(t)=e^{-t}+e^{-2t}$ you obtained is only valid for $t>0$. For $t<0$ the solution of the now homogeneous equation is $y(t)=2e^{-2t}$, so that in total $$ y(t)=2e^{-2t}+H(t)(e^{-t}-e^{-2t}). $$

Answer 2

Yes because in the definition of Laplace transform we can state:

$$L\{f(t) \}=\int_0^{\infty}e^{-st}f(t)dt=\int_{0^-}^{\infty}e^{-st}f(t)dt$$

Since with Impulse response and Heaviside response, our functions are not differentiable at $t=0$, so we take Laplace Transform from $t=0^-$, without loss of generality.

This not a rigorous explanation by any means, but it should show you why we chose $0^-$ instead of $0$