# Solutions of $y''+y'+by=\cos x$ that satisfy $\lim\limits_{x\to\infty}\frac{y_1(x)-y_2(x)}{e^x}=0$

Find for which values of $b$ every two solutions $y_1,y_2$ of the equation $y''+y'+by=\cos x$ satisfy $$\lim_{x\to\infty}\frac{y_1(x)-y_2(x)}{e^x}=0$$

My attempt:

Since the difference of every two solutions of the original equation is a solution of the homogeneous equation $y''+y'+by=0$ we are interested only in the solutions of the latter. The characteristic roots are:

$$m_{1,2}=\frac{-1\pm\sqrt{1-4b}}{2}$$

If $b=1/4$ then $m_{1,2}=-1/2$ so the general homogeneous solution is $y_h(x)=(C_1+C_2x)e^{-x/2}$ and $$\lim_{x\to\infty}\frac{(C_1+C_2x)e^{-x/2}}{e^x}=\lim_{x\to\infty}\frac{C_1+C_2x}{e^{3x/2}}=0$$

If $0 \leq b < \frac{1}{4}$ then $0<\sqrt{1-4b}\leq 1$ and so we have two distinct nonpositive real roots (at least one is strictly negative). Therefore the general homogeneous solution is $$y_h(x)=C_1 e^{m_1 x}+C_2e^{m_2 x}=\frac{C_1}{e^{|m_1|x}}+\frac{C_2}{e^{|m_2|x}}$$

So $$\lim_{x\to\infty}\frac{y_h(x)}{e^x}=\lim_{x\to\infty} \left( \frac{C_1}{e^{(|m_1|+1)x}}+\frac{C_2}{e^{(|m_2|+1)x}} \right )=0$$

If $b<0$ then $\sqrt{1-4b}>1$ we have two distinct real roots (with opposite signs). If $m_1<0$ and $m_2>0$ then the general homogeneous solution is $$y_h(x)=C_1 e^{m_1 x}+C_2e^{m_2 x}=\frac{C_1}{e^{|m_1|x}}+C_2e^{m_2 x}$$

Therefore

$$\lim_{x\to\infty}\frac{y_h(x)}{e^x}=\lim_{x\to\infty} \left( \frac{C_1}{e^{(|m_1|+1)x}}+C_2e^{(m_2-1) x} \right )=0 \iff m_2<1 \iff b>-2$$

If $b>\frac{1}{4}$ then there are two complex roots of the form $-\frac{1}{2}\pm i\beta$ so the general homogeneous solution is $$y_h(x)=e^{-x/2} (C_1 \cos (\beta x)+C_2 \sin (\beta x))$$

Hence

$$\lim_{x\to\infty}\frac{y_h(x)}{e^x}=\lim_{x\to\infty} \left( e^{-3x/2} (C_1 \cos (\beta x)+C_2 \sin (\beta x)) \right )=0$$

In conclusion for all $b>-2$ every two solutions $y_1,y_2$ of the equation $y''+y'+by=\cos x$ satisfy $$\lim_{x\to\infty}\frac{y_1(x)-y_2(x)}{e^x}=0$$

Is it correct? If yes, is there an easier way to solve this problem?

## 1 Answer

All possible values of the difference $y_1(x)-y_2(x)$ is the whole set of the homogeneous equation solutions. Therefore your question is equivalent to the following:

Find for which values of $b$ every solution $y$ of the equation $y''+y'+by=0$ satisfy $$\lim_{x\to\infty}\frac{y(x)}{e^x}=0$$ $(1)$

The homogeneous equation has two basis solutions $u_1(x), u_2(x)$ which depends on $b$. Then the basic inequalities considerations can be taken into account.

1. If $D = 1-4b = 0$ then we have $u_1(x) = e^{-x/2}, u_2(x)=xe^{-x/2}$ and (1) is satisfied.

2. If $D<0$ then the real part of basis solutions equals to $e^{-x/2}$ which is also good for you. This case gives $1-4b<0, b>1/4$.

3. In the last case $D>0$ we must demand the biggest of two (real) exponent to be less than $e^x$. Which finally leads us to

$\frac{-1+\sqrt{1-4b}}{2} < 1$

$b>-2$

Therefore, the answer is $b \in (-2, \infty)$