`y' = x(1+y)` Solve the differential equation

 An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows `(dy)/(dx)= f(x,y)` .

It can also be in a form of `N(y) dy= M(x) dx` as variable separable differential equation.

 To be able to set-up the problem as `N(y) dy= M(x) dx` , we let `y' = (dy)/(dx)` .

 The problem: `y'=x(1+y)` becomes:

`(dy)/(dx)=x(1+y)`

Rearrange by cross-multiplication, we get:

`(dy)/(1+y)=xdx`

Apply direct integrationon both sides:...

 An ordinary differential equation (ODE) has differential equation for a function with single variable. A first order ODE follows `(dy)/(dx)= f(x,y)` .

It can also be in a form of `N(y) dy= M(x) dx` as variable separable differential equation.

 To be able to set-up the problem as `N(y) dy= M(x) dx` , we let `y' = (dy)/(dx)` .

 The problem: `y'=x(1+y)` becomes:

`(dy)/(dx)=x(1+y)`

Rearrange by cross-multiplication, we get:

`(dy)/(1+y)=xdx`

Apply direct integration on both sides: `int (dy)/(1+y)= int xdx` to solve for the general solution of a differential equation.

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

`u= 1+y` then `du = dy`

The integral becomes:  `int(dy)/(1+y)=int(du)/(u)`

 Applying basic integration formula for logarithm:

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

Plug-in `u = 1+y` on `ln|u|` , we get:

`int(dy)/(1+y)=ln|1+y|`

For the right side, we apply the Power Rule of integration: `int x^n dx = x^(n+1)/(n+1)+C`

`int x* dx= x^(1+1)/(1+1)+C`

             ` = x^2/2+C`

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

`ln|1+y|= x^2/2+C`

or

`y =e^((x^2/2+C))-1`