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

Question

7009 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-08T15:25:46

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

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) #