Thread: Hi. I'm struggling with this problem. Someone helps me please. Thank you.

2. Re: Hi. I'm struggling with this problem. Someone helps me please. Thank you.

$a=0.2,~b=0.1$

$\dfrac {dV}{dt} = a V e^{-bt}$

$\dfrac{dV}{V} = a e^{-b t} ~dt$

$\ln(V) = -\dfrac a b e^{-b t} + C$

$V(t) = e^{-\frac a b e^{-b t}}e^C = e^C e^{-2e^{-0.1 t}}$

$V(0) = e^C e^{-2} = 1.86$

$e^C \approx 13.7436$

$V(t) \approx 13.7436 e^{-2e^{-0.1 t}}$

3. Re: Hi. I'm struggling with this problem. Someone helps me please. Thank you.

Another approach is to write the given ODE in standard linear form:

$\frac{dV}{dt}-0.2e^{-0.1t}V=0$

Compute the integrating factor:

$\mu(t)=\exp\left(-0.2\int e^{-0.1t}\,dt\right)=e^{2e^{-0.1t}}$

Multiply through by the integrating factor:

$e^{2e^{-0.1t}}\frac{dV}{dt}-0.2e^{-0.1t}e^{2e^{-0.1t}}V=0$

Observe that the LHS can be rewritten as the derivative of a product:

$\frac{d}{dt}\left(e^{2e^{-0.1t}}V\right)=0$

Integrate through with respect to $t$:

$e^{2e^{-0.1t}}V=C$

Thus:

$V(t)=Ce^{-2e^{-0.1t}}$

Use the initial conditions to find $C$

$V(0)=Ce^{-2e^{-0.1(0)}}=Ce^{-2}=1.86\implies C=1.86e^2$

And so the solution to the IVP is:

$V(t)=1.86e^2e^{-2e^{-0.1t}}=1.86e^{2-2e^{-0.1t}}$

This is equivalent to the solution posted by romsek, just written in a slightly different form.