differential equation?

Question

whats the particular integral of:

y'' - 3y' +2y = x^2 +3x -4

I used D-operators method and got the answer:
y_p= (x^2)/2 + 3x + 2.

whats the correct answer please?

Answer 1

Sinceyou have a diffrenetla equation of order 2 and the right hand side is a polynomila of degree 2, y_p = ) where a, b and c are constants. Then,

y'' - 3y' +2y = 2a - 3(2ax + b) + 2(ax^2 + bx + c) = 2ax^2 + (2b- 6a)x + 2a - 3b + 2c = x^2 + 3x - 4, So,

2a = 1 => a = 1/2
2b - 6a = 2b - 3 = 3 => b = 3
2a - 3b + 2c = 1 - 9 + 2c = - 4 => 2c = 4 => c = 2.

So, y_p = (x^2)/2 + 3x + 2, your answer is right!

Of course, we could come to this same conclusion much quickier djust plugging y _p= (x^2)/2 + 3x + 2in the differential equation.

Answer 2

If y(x) = (x^2)/2 + 3x + 2, then
y'(x) = x + 3, and y''(x) = 1, so
y'' - 3y' + 2y =
1 - 3(x+3) + 2 [ (x^2)/2 + 3x + 2 ] =
1 - 3x - 9 + (x^2) + 6x + 4 =
x^2 + 3x - 4

So it checks out. Given the nature of the function here, I wouldn't say you're missing a "+ C" or coefficients that could offer other general solutions.

Answer 3

Your answer is correct
y = (x^2)/2 + 3x + 2.
y' = x + 3
y" = 1

substitute:
1 - 3(x+3) + x^2 + 6x + 4 = x^2 + 3x -4