Question

What is the solution of the differential equation? : `y'-2xy = 1 \ \` with `\ \ y(0)=y_0`

Answer 1

`y = (sqrt(pi/2)erf(x) + y_0)e^(x^2)`

Explanation:

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

`dy/dx + P(x)y=Q(x)`

We have:

`y'-2xy = 1 \ \` with `\ \ y(0)=y_0` ..... [1]

This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, `I`, using;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ -2x \ dx)`
`\ \ = exp( -x^2 )`
`\ \ = e^(-x^2)`

And if we multiply the DE [1] by this Integrating Factor, `I`, we will have a perfect product differential;

`dy/dxe^(-x^2) - 2xye^(-x^2) = e^(-x^2)`

`:. d/dx( ye^(-x^2))= e^(-x^2)`

This has transformed our initial ODE into a Separable ODE, so we can now "separate the variables" to get::

`ye^(-x^2) = int \ e^(-x^2) \ dx + C`

We now encounter a problem, as the RHS integral is not one that can be evaluated using standard elementary functions, and so we introduce the error function (or Gauss error function ):

`erf(x) = 2/sqrt(pi) int_0^x \ e^(-t^2) \ dt`

From which we get:

`ye^(-x^2) = sqrt(pi/2)erf(x) + C`
`:. y = (sqrt(pi/2)erf(x) + C)e^(x^2)`

And, if we apply the initial condition, `y(0)=y_0` we get:

`y_0 = (sqrt(pi/2)) xx 0 + C => C=y_0`

Leading to the Specific Solution:

`:. y = (sqrt(pi/2)erf(x) + y_0)e^(x^2)`