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

1 Answer
May 13, 2018

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

Explanation:

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 ))