Need help with substitution for integration

Question

I'm trying to go through the solution for a SIS epidemic model, but I'm stuck figuring out the following steps.

So we have

$i′=r(k−i)i−αi$

And to solve this, they did a substitution $y=i^{−1}$. They integrated and came up with

$ye^{(rk−α)t}=∫re^{(rk−α)t}dt+C$

I have no idea how they ended up with this equation. So taking substitution $y = i^{-1}$, we get $\frac{dy}{dt} = -\frac{1}{i^2}$, but i don't know how to use that for the first equation.

Did they use separation of variables to solve this after the substitution? Anyone who could show me how they got to the second equation would be appreciated.

Thank you!

Answer 1

Regroup $$i′=r(k−i)i−αi = (rk-\alpha)i-ri^2$$

Divide both sides by $i^2$ to get $$i^{-2}i' =(rk-\alpha)i^{-1} -r$$

Let $$y=i^{-1}$$

We get $$ -y'=(rk-\alpha)y -r$$

This is a linear equation with the given solution.

Answer 2

From $ -\mathrm{y}^{\prime}=(\mathrm{rk}-\alpha) \mathrm{y}-\mathrm{r} $, we get $\mathrm{y}^{\prime}=-(\mathrm{rk}-\alpha) \mathrm{y}+\mathrm{r}$.

Thus we obtain the following linear first - order differential equation $$ \mathrm{y}^{\prime}+(\mathrm{rk}-\alpha) \mathrm{y}=\mathrm{r}.$$ We get that the integrating factor is $$ \mu(\mathrm{t})=\mathrm{e}^{\int(\mathrm{rk}-\alpha) \mathrm{dt}}=\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}}. $$

Multiplying the differential equation by the integrating factor $\mu(\mathrm{t})$, we obtain $$\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}} \mathrm{y}^{\prime}+\mathrm{e}^{(\mathrm{rk}-\alpha) \mathrm{t}}(\mathrm{r} \mathrm{k}-\alpha) \mathrm{y}=\mathrm{re}^{(\mathrm{rk}-\alpha) \mathrm{t}},$$

which we can write as $$\left(e^{(\mathrm{rk}-\alpha) \mathrm{t}} \mathrm{y}(\mathrm{t})\right)^{\prime}=\mathrm{re}^{(\mathrm{rk}-\alpha) \mathrm{t}}.$$

Integrating both sides with respect to $t$,

$$ y e^{(r k-\alpha) t} = \int r e^{(r k-\alpha) t} d t+C $$