What is the particular solution of the differential equation? : y'+4xy=e^(-2x^2) with y(0)=-4.3

1 Answer
Sep 8, 2017

y = (x-4.3)e^(-2x^2)

Explanation:

We have:

y' +4xy = e^(-2x^2) ..... [A]

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)

The given equation is already in standard form, so the integrating factor is given by;

I = e^(int P(x) dx)
\ \ = exp(int \ 4x \ dx)
\ \ = exp( 2x^2 )
\ \ = e^(2x^2)

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

2e^(2x^2)y' +8xe^(2x^2)y = 2e^(2x^2)e^(-2x^2)

:. d/dx (2e^(2x^2)y ) = 2

Which we can now "seperate the variables" to get:

2e^(2x^2)y = int \ 2 \ dx

Which is trivial to integrate giving the General Solution:

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

Applying the initial condition we get:

2e^(0)(-4.3) = 0 + C => C = -8.6

Giving the Particular Solution:

2e^(2x^2)y = 2x -8.6
:. e^(2x^2)y = x -4.3
:. e^(2x^2)e^(-2x^2)y = (x-4.3)e^(-2x^2)
:. y = (x-4.3)e^(-2x^2)