Recall that in solving simple first order "ordinary differential equation" (ODE), we may apply **variable separable differential equation** wherein:

`N(y)y'=M(x)`

`N(y)(dy)/(dx)=M(x)`

`N(y) dy=M(x) dx`

Before we can work on the direct integration:` ` int N(y) dy= int M(x) dx to solve for the general solution of a differential equation.

For the given...

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Recall that in solving simple first order "ordinary differential equation" (ODE), we may apply **variable separable differential equation** wherein:

`N(y)y'=M(x)`

`N(y)(dy)/(dx)=M(x)`

`N(y) dy=M(x) dx`

Before we can work on the direct integration:` ` int N(y) dy= int M(x) dx to solve for the general solution of a differential equation.

For the given first order ODE: `(dy)/(dx)=y+3 ` can be rearrange by cross-multiplication into:

`(dy)/(y+3)=dx`

Apply direct integration on both sides:` int(dy)/(y+3)=int ` dx

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

`u= y+3` then` du = dy`

The integral becomes:

`int(dy)/(y+3)=int(du)/(u)`

Applying basic integration formula for logarithm:

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

Plug-in `u = y+3` on `ln|u`` |` , we get:

`int(dy)/(y+3)=ln|y+3|`

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|y+3|= x+C`

or

`y =e^(x+C)-3`

`y =Ce^x-3`