ISI2015-MMA-89

Question

Let $y(x)$ be a non-trivial solution of the second order linear differential equation $$\frac{d^2y}{dx^2}+2c\frac{dy}{dx}+ky=0,$$ where $c<0$, $k>0$ and $c^2>k$. Then

• A. $\mid y(x) \mid \to \infty$ as $x \to \infty$
• B. $\mid y(x) \mid \to 0$ as $x \to \infty$
• C. $\underset{x \to \pm \infty}{\lim} \mid y(x) \mid$ exists and is finite
• D. none of the above is true

Comments

B)

1.  As $c<0$ put $c_1=-c$. 
2. Find the roots of the auxiliary equation.
3. You will get real and distinct roots as $c^2>k$.
4. Take $\lim_{x\rightarrow \infty} y(x) $.

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As both the roots of the auxiliary equation are positive, so $y(x)\rightarrow +- infinity$ (according to the sign of the arbitrary constant) as x tends to infinity.

Option a) should be correct

Answer 1

@Idon'tknow

We will get exp(-cx) as a common term in our solution. Therefore solution tends to zero as x tends to infinity.

Comments

Since, $c<0$, $e^{-cx}$ goes to infinity as $x$ goes to infinity.

Precisely, here non-trivial solution is, $y(x) = e^{\sqrt{c^2-k}x-cx}\left(\frac{c_1}{e^{2\sqrt{c^2-k}x}}+c_2\right)$ for some arbitrary constants $c_1$ and $c_2.$

So, $|y(x)| \rightarrow \infty $ as $x \rightarrow \infty$