What is a solution to the differential equation? : #dy/dx + (1-2x)/x^2*y = 1# when #y|_(x=1)=0#

We have:

# dy/dx + (1-2x)/x^2y = 1# and #y(1)=0#

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) #

So we compute and integrating factor, #I#, using;

# I = e^(int P(x) dx) #
# \ \ = exp(int \ (1-2x)/x^2 \ dx) #
# \ \ = exp( -2lnx -1/x ) #
# \ \ = e^( -lnx^2 -1/x ) #
# \ \ = e^( -lnx^2)e^( -1/x ) #
# \ \ = 1/x^2e^( -1/x ) #

And if we multiply the original DE by this Integrating Factor, #I#, we will have (by design) a perfect product differential;

# 1/x^2e^( -1/x )dy/dx + 1/x^2e^( -1/x ) (1-2x)/x^2y = 1/x^2e^( -1/x ) #

# :. d/dx{ 1/x^2e^( -1/x ) y } = 1/x^2e^( -1/x ) #

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

# 1/x^2e^( -1/x ) y = int \ 1/x^2e^( -1/x ) \ dx#

We can integrate, and we get:

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

We now apply the initial conditions, #y(1)=0# to gte:

# 0 = e^( -1 ) + C => C = -e^( -1 )#

Leading to the Particular Solution is:

# 1/x^2e^( -1/x ) y = e^( -1/x ) -e^( -1 ) #

# :. y = x^2 - x^2e^( 1/x )e^( -1 ) #

# :. y = x^2 - x^2e^( 1/x-1 ) #

# :. y = x^2(1 - e^( 1/x-1 )) #