The usual approach for an equation like this is to rewrite slightly, and then find an integrating factor:
$$y' + \frac 2 t y = \frac{\sin t}{t}$$
We want to find a function $\mu$ so that the left hand side can be written as a single derivative; to this end, multiply through by $\mu$ to get
$$\mu y' + \frac {2 \mu}{t} y= \mu \frac{\sin t}{t}$$
Now notice that the left hand side looks like product rule, if only we could write
$$\mu y' + \frac{2\mu}{t} y = \mu y' + \mu' y$$
So if we can solve
$$\mu' = \frac{2\mu}{t}$$
we'll be a lot closer; but this can be solved using
$$\frac{2}{t} = \frac{\mu'}{\mu} = \left(\ln \mu\right)'$$
Integrating, and choosing the constant of integration so that $\mu(0) = 1$, this leads to
$$\ln \mu = 2 \ln t \implies \boxed{\mu = t^2}$$
So (provided $t \ne 0$) we can rewrite the original equation as
$$ (t^2 y)' = t^2 y' + 2t y = t \sin t$$
Now integrate and solve for $y$.
Note that $S+I$ is a constant $K$; you can see that by adding the equations. Then the first equation becomes
$$\frac{dS}{dt} = -S (K-S) \frac{B}{N} $$
This equation is separable as follows:
$$\frac{dS}{S (K-S)} = -\frac{B}{N} dt $$
which is equivalent to
$$dS \frac{1}{K} \left ( \frac1{S} + \frac1{K-S} \right ) = -\frac{B}{N} dt$$
Now integrate; hopefully you see this is easy:
$$\log{\left ( \frac{S}{K-S} \right )} = C - K \frac{B}{N} t $$
where $C$ is a constant of integration. Solve for $S$. To get $I$, note that $I = K-S$.
You may find $K$ and $C$ by proper initial conditions on $S$ and $I$.
Best Answer
Well, first of all we have:
$$\mathcal{H}'\left(t\right)=\text{a}\cdot\left\{\frac{\mathcal{H}\left(t\right)}{\mathcal{H}\left(t\right)+\text{b}}-\text{p}\right\}\tag1$$
We can write (by dividing both sides by $\frac{\mathcal{H}\left(t\right)}{\mathcal{H}\left(t\right)+\text{b}}-\text{p}$):
$$\frac{\mathcal{H}'\left(t\right)}{\frac{\mathcal{H}\left(t\right)}{\mathcal{H}\left(t\right)+\text{b}}-\text{p}}=\text{a}\space\Longrightarrow\space\int\frac{\mathcal{H}'\left(t\right)}{\frac{\mathcal{H}\left(t\right)}{\mathcal{H}\left(t\right)+\text{b}}-\text{p}}\space\text{d}t=\int\text{a}\space\text{d}t\tag2$$
Which gives for the RHS:
$$\int\text{a}\space\text{d}t=\text{a}\cdot t+\text{K}_1\tag3$$
And for the LHS, we substitute $\text{u}=\mathcal{H}\left(t\right)$:
$$\int\frac{\mathcal{H}'\left(t\right)}{\frac{\mathcal{H}\left(t\right)}{\mathcal{H}\left(t\right)+\text{b}}-\text{p}}\space\text{d}t=\int\frac{1}{\frac{\text{u}}{\text{u}+\text{b}}-\text{p}}\space\text{d}\text{u}\tag4$$
Now, for the integral we get:
$$\int\frac{1}{\frac{\text{u}}{\text{u}+\text{b}}-\text{p}}\space\text{d}\text{u}=\frac{\text{b}\cdot\ln\left|\text{b}\cdot\text{p}+\text{u}\cdot\left(\text{p}-1\right)\right|}{\left(\text{p}-1\right)^2}-\frac{\text{u}}{\text{p}-1}+\text{K}_2\tag5$$
So, in the end we have:
$$\frac{\text{b}\cdot\ln\left|\text{b}\cdot\text{p}+\mathcal{H}\left(t\right)\cdot\left(\text{p}-1\right)\right|}{\left(\text{p}-1\right)^2}-\frac{\mathcal{H}\left(t\right)}{\text{p}-1}=\text{a}\cdot t+\text{K}\tag6$$