Question

How do you solve the separable differential equation `dy/dx=(cosx)e^(y+sinx)`?

Answer 1

`y = ln(1/(A-e^sinx))` is the General Solution

Explanation:

We have:

`dy/dx = (cosx)e^(y+sinx)`
`dy/dx = (cosx)e^ye^sinx`

Which is a First Order Separable Differential Equation, which we can rewrite as:

`1/e^ydy/dx = (cosx)e^sinx`

We can then "separate the variables" to get:

`int \ e^-y \ dy = int \ (cosx)e^sinx \ dx`

Which we can directly (and easily) integrate to get:

`- e^-y = e^sinx + B`

`:. e^-y = A-e^sinx`

`:. -y = ln(A-e^sinx)`

`:. y = -ln(A-e^sinx)`

`:. y = ln(1/(A-e^sinx))`

Answer 2

`y=-lnabs(C-e^sinx)`

Explanation:

To "separate" a differential equation means to move all the terms with `y` to one side of the equation, and all the terms with `x` to the other side of the equation.

We treat `dy/dx` as a division problem, so we can move the `dx` to the other side of the equation, leaving just `dy`.

To separate this, we also need to split up `e^(y+sinx)` as `e^y(e^sinx)`.

`dy/dx=(cosx)e^(y+sinx)`

`dy=(cosx)e^y(e^sinx)dx`

`dy/e^y=(cosx)e^sinxdx`

Now we can integrate both sides:

`inte^-ydy=inte^sinx(cosx)dx`

For the left-hand side, use the substitution `u=-y`, implying that `du=-dy`.

`-inte^-y(-dy)=inte^sinxcosxdx`

`-inte^u=inte^sinxcosxdx`

`-e^u=inte^sinxcosxdx`

`-e^-y=inte^sinxcosxdx`

Following a similar process on the right, let `t=sinx` so `dx=cosxdx`.

`-e^-y=inte^tdt`

`-e^-y=e^sinx+C`

Solving for `y`:

`e^-y=-e^sinx+C`

`-y=lnabs(C-e^sinx)`

`y=-lnabs(C-e^sinx)`