# `dy/dx = xe^(x^2)` Use integration to find a general solution to the differential equation

An ordinary differential equation (ODE)  is differential equation for the derivative of a function of one variable. When an ODE is in a form of `y'=f(x,y)` , this is just a first order ordinary differential equation.

We may express `y' ` as `(dy)/(dx) ` to write in a form of...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

An ordinary differential equation (ODE)  is differential equation for the derivative of a function of one variable. When an ODE is in a form of `y'=f(x,y)` , this is just a first order ordinary differential equation.

We may express `y' ` as `(dy)/(dx) ` to write in a form of `(dy)/(dx)= f(x,y)` and apply variable separable differential equation: `N(y)dy = M(x) dx` .

The given problem: ` (dy)/(dx) = xe^(x^2) `  can be rearrange as:

`dy= xe^(x^2) dx`

Apply direct integration on both sides:

`int dy= int xe^(x^2) dx`

For the left side, we apply basic integration property: `int (dy)=y` .

For the right side, we may apply u-substitution by letting: `u = x^2` then `du =2x dx` or  `(du)/2=x dx` .

The integral becomes:

`int xe^(x^2) dx=int e^(x^2) *xdx`

` =int e^(u) *(du)/2`

Apply the basic integration property: `int c*f(x)dx= c int f(x) dx` .

`int e^(u) *(du)/2=(1/2)int e^(u) du`

Apply basic integration formula for exponential function:

`(1/2)int e^(u) du=(1/2)e^(u)+C`

Plug-in `u=x^2` on `(1/2)e^(u)+C` , we get:

`int xe^(x^2) dx=(1/2)e^(x^2)+C`

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

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

or

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

Approved by eNotes Editorial Team