harold domar growth model: solve the differential equation by integration

Question

harold domar growth model: solve the differential equation (by integration)

When arranging the differential equation like this:

$$ \frac{dY}{dt} \cdot \frac{1}{Y(t)} = (\frac{s}{v})$$ and integrating both parts, I get the solution: $Y(t) = e^((\frac{s}{v})\cdot)\cdot C$, which is correct apparently

$C$ is the intergration constant, $s$ and $v$ are some parameters of the Harold Domar model

However, whenever I try to use the same technique of integrating right and left hand side when rearranging the equation like this:

$\frac{dY}{dt} - \frac{s}{v}\cdot Y(t) = 0$ , I get nowhere, well, I get something like $Y(1-\frac{s}{v} \cdot t) = c$.

How come it doesn't seem to work with different rearrangements of the equation? Does it matter? Or is it just my lack of integration skills? Many other rearrangements are possible, but I only find the solution with this particular one and I wonder why.

Answer 1

The problem with your second approach is that you do not know what the actual expression for $y(t)$ is before hand, so integrating is not possible. Here is what I mean, the original expression is

$$ \frac{{\rm d}y(t)}{{\rm d}t} - \frac{s}{v}y(t) = 0 $$

Which leads to

$$ \color{blue}{\int {\rm d}y} - \color{red}{\frac{s}{v}\int y(t)~{\rm d}t} = C $$

the $\color{blue}{\text{blue}}$ term is easy to get, but the $\color{red}{\text{red}}$ is not, since you don't know what the expression for $y = y(t)$ is.

Answer 2

$$Y'-\frac1\tau Y=0$$ can be integrated by noting that

$$\left(Y'-\frac1\tau Y\right)e^{-t/\tau}=Y'e^{-t\tau}+(e^{-t/\tau})'Y=(Ye^{-t/\tau})'.$$

Then

$$Ye^{-t/\tau}=C$$ and $$Y=Ce^{t/\tau}.$$

This technique brings no benefit as regards an homogeneous equation, but is useful to find non-homogeneous solutions.

$$Y'-\frac1\tau Y=Z$$ yields

$$Ye^{-t/\tau}=\int Z\,e^{-t/\tau}\,dt$$

and

$$Y=e^{t/\tau}\int Z\,e^{-t/\tau}\,dt.$$