Apply direct integration both sides: `intN(y) dy= int M(x) dx` to solve for the general solution of a differential equation.

For the given first order ODE: `(dy)/(dx)=6-y` it can be rearrange by cross-multiplication into:

`(dy)/(6-y)=dx`

Apply direct integration on both sides: `int(dy)/(6-y)=int dx`

For the left side, we consider u-substitution by letting:

`u=6-y ` then ` du = -dy` or `-du=dy`

The integral becomes: `int(dy)/(6-y)=int(-du)/(u)`

Applying basic integration formula for logarithm:

`int(-du)/(u)= -ln|u|`

Plug-in `u = 6-y` on ` -ln|u|` , we get:

`int(dy)/(6-y)=-ln|6-y|`

For the right side, we apply the basic integration: `int dx= x+C`

Combing the results from both sides, we get the **general solution of the differential equation** as:

`-ln|6-y|= x+C`

`y =6-e^((-x-C))`

or

`y = 6-Ce^(-x)

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