Question

How do you find all solutions of the differential equation `dy/dx=B+ky`?

Answer 1

`y=(Ce^(kx)-B)/k`

Explanation:

We have to first separate the variables, treating `dy/dx` as division and getting all `y` terms on one side and all `x` terms on the other.

First, we see that there are no `x` terms other than `dx`, so this will need to be on a side unto itself.

Further, we can't subtract `ky` to get `dy` and `ky` on the same side: it doesn't deal with the issues of multiplying later by `dx` and `dy` must be attached to all other `y` terms by multiplication.

So, what we can do is divide both sides by `B+ky` entirely. If we also multiply by `dx`, this should give us:

`dy/dx=B+ky`

`dy/(B+ky)=dx`

We then integrate:

`intdy/(B+ky)=intdx`

The left-hand integral can either be solved using the substitution `u=B+ky` or just by inspection. Remember `B` and `k` are constants (we'll also add a constant of integration).

`1/kint((k)dy)/(B+ky)=x+C`

`1/klnabs(B+ky)=x+C`

Multiply both sides by `k`. Even though `C` changes, we'll leave it as `C` since it's a catch-all for any constant.

`lnabs(B+ky)=kx+C`

`abs(B+ky)=e^(kx+C)`

Note that `e^(kx+C)=e^(kx)e^C=Ce^(kx)` where `C` again represents `e^C`, some other constant.

`abs(B+ky)=Ce^(kx)`

The absolute value bars are just a reminder that it's possible that `B+ky<0`.

`ky=Ce^(kx)-B`

`y=(Ce^(kx)-B)/k`