Question

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

Answer 1

`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 ))`