General solution of linear differential equation y '- 2Y = - 2x + 3

General solution of linear differential equation y '- 2Y = - 2x + 3

The type is y '+ P (x) y = q (x). P (x) = - 2, q (x) = - 2x + 3,
-2X is a primitive function of P (x)
Then find ∫ Q (x) e ^ (- 2x) DX = ∫ (- 2x + 3) e ^ (- 2x) DX = ∫ (- 2x) e ^ (- 2x) DX + 3 ∫ e ^ (- 2x) DX = [∫ (- 2x) e ^ (- 2x) DX] - (3 / 2) e ^ (- 2x)
And the first of them
∫(-2x)e^(-2x)dx=-(1/2）∫(-2x)e^(-2x)d(-2x)=-(1/2）∫(-2x)d(e^(-2x))
=-(1 / 2) [(- 2x) e ^ (- 2x) - ∫ (e ^ (- 2x) d (- 2x)] = - (1 / 2) [(- 2x) e ^ (- 2x) - (e ^ (- 2x)] + C0 (C0 is any constant)
So ∫ Q (x) e ^ (- 2x) DX = Xe ^ (- 2x) - e ^ (- 2x) + C1, (C1 is an arbitrary constant)
From the formula of the solution, it is concluded that:
Y = e ^ (2x) [C + ∫ Q (x) e ^ (- 2x) DX] = e ^ (2x) [C + Xe ^ (- 2x) - e ^ (- 2x)], (C is any constant)
So the general solution is y = CE ^ (2x) + X-1, (C is any constant)
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(put it into the equation and verify it.)