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