$a=0.2,~b=0.1$
$\dfrac {dV}{dt} = a V e^{-bt}$
$\dfrac{dV}{V} = a e^{-b t} ~dt$
$\ln(V) = -\dfrac a b e^{-b t} + C$
$V(t) = e^{-\frac a b e^{-b t}}e^C = e^C e^{-2e^{-0.1 t}}$
$V(0) = e^C e^{-2} = 1.86$
$e^C \approx 13.7436$
$V(t) \approx 13.7436 e^{-2e^{-0.1 t}}$
Another approach is to write the given ODE in standard linear form:
Compute the integrating factor:
Multiply through by the integrating factor:
Observe that the LHS can be rewritten as the derivative of a product:
Integrate through with respect to :
Thus:
Use the initial conditions to find
And so the solution to the IVP is:
This is equivalent to the solution posted by romsek, just written in a slightly different form.