Question

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Every solution of the differential equation $y'' + 2y' + 2y = 0$ tends to zero as $x \to \infty$.
Reason R : The real part of the roots of the polynomial $\lambda^2 + 2\lambda + 2$ are negative.
In the light of the above statements, choose the correct answer from the options given below

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is NOT the correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Hint:

Stability Condition: In $y'' + a y' + b y = 0$, every solution $\to 0$ as $x \to \infty$ if and only if $a > 0$ and $b > 0$ (Routh-Hurwitz criterion for order 2). Here $a = 2 > 0$ and $b = 2 > 0$, confirming asymptotic stability!

Answer 1

The Correct Option is A

Step 1: Concept
For a linear homogeneous differential equation with constant coefficients $a y'' + b y' + c y = 0$, all solutions decay to zero as $x \to \infty$ (asymptotically stable) if and only if all roots $\lambda$ of the characteristic polynomial $a\lambda^2 + b\lambda + c = 0$ have strictly negative real parts ($\text{Re}(\lambda) < 0$).

Step 2: Key Formulas and Approach
Solve the characteristic equation $\lambda^2 + 2\lambda + 2 = 0$ using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Analyze the behavior of $e^{\lambda x} = e^{\text{Re}(\lambda) x} e^{i \text{Im}(\lambda) x}$ as $x \to \infty$.

Step 3: Step-by-step Explanation

• Find the roots of the characteristic polynomial $\lambda^2 + 2\lambda + 2 = 0$: \[ \lambda = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i \]
• The real part of both complex roots is $\text{Re}(\lambda) = -1 < 0$. Hence, Reason R is true.
• Construct the general solution of the differential equation: \[ y(x) = e^{-x} \left( C_1 \cos x + C_2 \sin x \right) \]
• Evaluate the limit as $x \to \infty$: \[ \lim_{x \to \infty} y(x) = \lim_{x \to \infty} e^{-x} (C_1 \cos x + C_2 \sin x) = 0 \] because $e^{-x} \to 0$ as $x \to \infty$, while $(C_1 \cos x + C_2 \sin x)$ remains bounded. Hence, Assertion A is true.
• Since the decaying exponential factor $e^{\text{Re}(\lambda)x} = e^{-x}$ directly causes every solution to approach zero, Reason R provides the precise mathematical explanation for Assertion A.

Step 4: Final Answer
Both A and R are true and R is the correct explanation of A. Thus, Option (A) is correct.