Question

Question #e4bc9

Answer 1

See below.

Explanation:

Calling `z(x)=y(x)+x` and substituting `y(x) = z(x)-x` into the differential equation we get at

`z'(cosz+2)=4` or

`(cosz+2)dz = 4 dx`

This is a separable differential equation. Now integrating both sides

`sinz+2z = 4x+C` or

`sin(y(x)+y)+2(y(x)+z)-4x=C`

which is a solution in implicit form.

Attached a plot for

`sin(y(x)+y)+2(y(x)+z)-4x=0`

Answer 2

`y' = (2 - cos(y+x))/(2+cos(y+x))`.

Explanation:

Solve for `y'`

`(y'+1)cos(y+x) = 2-2y'`

Open the parentheses by distributing `cos(y+x)`.

`y'cos(y+x) +cos(y+x) = 2-2y'`.

Collect terms tat include a factor of `y'` on one side (I'll use the left) and terms without `y'` on the other side of the equality.

`y'cos(y+x) + 2y' = 2 - cos(y+x)`.

Factor out the `y'`

`y'(cos(y+x) + 2) = 2 - cos(y+x)`.

Divide to isolate `y'`

`(y'(cos(y+x) + 2))/(cos(y+x) + 2) = (2 - cos(y+x))/(cos(y+x) + 2)`.

Simplify

`y' = (2 - cos(y+x))/(cos(y+x) + 2)`.

You may prefer to write

`y' = (2 - cos(y+x))/(2+cos(y+x))`.

Notation

The `cos(y+x)` makes this look more intimidating than it needs to be.

Let's replace `cos(y+x)` by `C`

We start with

`(y'+1)C= 2-2y'`

and we want to find `y'`. We proceed.

`y'C+C=2-2y'`

`y'C+2y' = 2-C`

`y'(C+2) = 2-C`

`y' = (2-C)/(C+2) = (2-C)/(2+C)`