JSM Learn - Q-70. Which of the following is the general solution of 2 2 2 0d y dy y dx dx − + =? (A) y = (Ax + B)ex      (B) y = (Ax + B)e–x     (C) y = Ae x + Be–x     (D) y = Acosx + Bsinx

Q-70. Which of the following is the general solution of 2 2 2 0d y dy y dx dx − + =?

(A) y = (Ax + B)e^x (B) y = (Ax + B)e^–x (C) y = Ae x + Be^–x (D) y = Acosx + Bsinx

Ans:(A)

Explanation:[Method - I]

Given that, $\frac{d^{2} y}{{dx}^{2}} - 2 \frac{dy}{dx} + y = 0 \Rightarrow D^{2} y - Dy + y = 0, Where D = \frac{d}{dx} \Rightarrow [D^{2} - D + 1] y = 0 The auxiliary equation is m^{2} - 2 m + 1 = 0 {(m - 1)}^{2} = 0 \Rightarrow m = 1, 1 Since, the roots are real and equal . ∴ CF = [Ax + B] e^{x} \Rightarrow y = [Ax + B] e^{x} [∵ Since, its roots of auxiliary equation are real and equal say (m), then [C . F = [c, x + c_{2}] e^{mx}]]$

$Method - II : [Option Verification] For y = [Ax + B] e^{x} \Rightarrow \frac{dy}{dx} = [Ax + B] e^{x} + {Ae}^{x} = [Ax + A + B] e^{x} \Rightarrow \frac{d^{2} y}{{dx}^{2}} = [Ax + A + B] e^{x} + {Ae}^{x} = [Ax + 2 A + B] e^{x} \Rightarrow \frac{d^{2} y}{{dx}^{2}} - 2 (\frac{dy}{dx}) + y \Rightarrow [Ax + 2 A + B] e^{x} - 2 [Ax + A + B] e^{x} + [Ax + B] e^{x} \Rightarrow 0$

Q-70. Which of the following is the general solution of 2 2 2 0d y dy y dx dx − + =?

(A) y = (Ax + B)e^x (B) y = (Ax + B)e^–x (C) y = Ae x + Be^–x (D) y = Acosx + Bsinx

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