solve the first order linear differential equation? (x-1)y' + y = x^2 - 1?

Answer 1

(x-1) y' + y = d/dx ((x-1) y), so we get
(x-1) y = ∫(x^2 - 1) dx
= x^3 / 3 - x + c
So y = x^3 / (3(x-1)) - x / (x-1) + c / (x-1)
= x^3 / (3(x-1)) - 1 + c / (x-1)
where we write x / (x-1) as (x-1 + 1) / (x-1) = 1 + 1/(x-1) and absorb the 1/(x-1) into the c/(x-1) term.

Answer 2

This differential equation has a polynomial solution. Since the right hand side is quadratic we can assume y is of the form

y = ax^2 + bx + c

So y' = 2ax + b. Plugging into the differential equation

(x-1)(2ax + b) + (ax^2 + bx + c) = x^2 -1

Now we need some carpentry...

(2ax^2 - 2ax + bx - b) + (ax^2 +bx + c) = x^2 -1

So

3ax^2 +2(b - a)x + (c - b) = x^2 -1.

Therefore 3a = 1, so a = 1/3 (we are equating coefficients of like powers of x). b - a must be 0, since there is no x term in the right hand side. so b = a = 1/3. Finally c - b = -1 implies
c = -1 +1/3 = -2/3. Plug the coefficients into the proposed solution and we get:

y = (1/3)x^2 + (1/3)x - 2/3 as the solution. We can clean this up a little bit and get

y = (x^2 + x - 2)/3 as the solution to the ODE.