# 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 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

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