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I am stuck on the following question:

Assuming $\frac{d^2y}{dx^2}-q(x)y = 0,\;\; 0 \le x \lt \infty ,\;\;y(0)=1,\;\;y'(0)=1$ wherein $q(x)$ is monotonically increasing continuous function,then which one of the following is true.

(a) $y(x) \to \infty$ as $x \to \infty$

(b) $\frac{dy}{dx}\to \infty$ as $x \to \infty$

(c) $y(x)$ has finitely many zeros in $[0,\infty)$

(d) $y(x)$ has infinitely many zeros in $[0,\infty)$

I am completely stuck on it.can anyone help me please.

  • 0
    @Babak Sorouh: For better results when searching, I would in general not use any abbreviations.2012-12-17

1 Answers 1

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Given $\frac{d^2y}{dx^2}-q(x)y = 0,\;\; 0 \le x \lt \infty ,\;\;y(0)=1,\;\;y'(0)=1$, where q(x) is positive monotonically increasing continuous functions .

Let q(x)=1, then we have $\frac{d^2y}{dx^2}-y = 0 \Rightarrow \lambda^2-1=0\Rightarrow \lambda= \pm1$

$\therefore CF=y(x)=c_1e^x+c_2e^{-x}\\ \frac{dy}{dx}(x)=c_1e^x-c_2e^{-x} $

Now , y(0)=1

$\Rightarrow 1=c_1+c_2$ and $\frac{dy}{dx}(0)=1 \Rightarrow 1=c_1-c_2 \Rightarrow c_1=1,c_2=0$

Hence $y(x)=e^x \Rightarrow \lim_{x\rightarrow \infty} y(x)=\lim_{x\rightarrow \infty}=\infty$

and $\frac{dy}{dx}=e^x \Rightarrow \lim_{x\rightarrow \infty} \frac{dy}{dx}=\lim_{x-\rightarrow \infty} e^x=\infty$

Hence, option (c) and (d) are incorrect

Therefore , option (a) and (b) are correct

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    B is also incorrect, take $q\equiv 0$, then $y'=1\not\to\infty$.2016-12-10