# Show that every solution tends to zero as $x→∞$ if $a_1>0$ for $y''+a_1y'+a_2y=0$.

Consider the equation $$y'' + a_1y' + a_2y = 0$$, where the constants $$a_1, a_2$$ are real. Suppose $$α + iβ$$ is a complex root of the characteristic polynomial, where $$α, β$$ are real, $$β≠0$$.

Show that every solution tends to zero as $$x→∞$$ if $$a_1>0$$.

My solution:

$$y'' + a_1y' + a_2y = 0$$

$$r^2 + a_1r + a_2 = 0$$

$$x = \frac{-a_1}{2} ±\frac{\sqrt{a_1^2-4a_2}}{2}$$

$$\varphi(x) = c_1e^{(\frac{-a_1}{2}+\frac{\sqrt{a_1^2-4a_2}}{2})x} + c_2e^{(\frac{-a_1}{2}-\frac{\sqrt{a_1^2-4a_2}}{2})x}$$

• Now you write out your solution in terms of exponentials and evaluate the limit Feb 19, 2017 at 16:47
• To further expand on my comment, you also know that there is an imaginary root (and thus there are two since they come in conjugate pairs) this means the radical is imaginary which leads to sines/cosines. So the most important part of the solution in terms of long term behavior is the $e^{-\frac{a_1 x}{2}}$ part Feb 19, 2017 at 17:18
• @socrates, are you and lovesTrumpsHate one and the same person? I suggest that you give me an honest answer and spare the moderators the pain of an investigation. Feb 19, 2017 at 21:16
• @AlexM. No, we are not. I know why you came to that conclusion. His questions are very much related to my questions. I do not know him. Feb 19, 2017 at 21:44

If $a_1^2<4a_2$, then $\sqrt{a_1^2-4a_2}=i\sqrt{4a_2-a_1^2}$ is imaginary and

\begin{align} \lim_{x\to \infty}e^{\frac12\left(-a_1\pm\sqrt{a_1^2-4a_2}\right)x}&=\lim_{x\to \infty}e^{-a_1x/2}e^{\pm\frac i2\sqrt{4a_2-a_1^2}}\\\\ &=\lim_{x\to \infty}e^{-a_1x/2}\left(\cos\left(\frac12\sqrt{4a_2-a_1^2}x\right)\pm i\sin\left(\frac12\sqrt{4a_2-a_1^2}\right)x\right)\\\\ &= 0 \end{align}

If $a_1^2\ge 4a_2$, and $a_2>0$, then $0\le \sqrt{a_1^2-4a_2}<a_1$.

Hence $-a_1+\sqrt{a_1^2-4a_2}< 0$ and $e^{\frac12\left(-a_1\pm\sqrt{a_1^2-4a_2}\right)x}\to 0$ as $x\to \infty$.

• Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark Feb 19, 2017 at 20:17
• Part 2 is true but not in the Q, which posits that the char. polynomial has a non-real root. For Part 1 you could point out that if there is a non-real root then there are exactly 2 roots, neither real, and that it must be true that $a_1^2<4a_2.$ (Note the different role of the "if"), and the rest follows...... You could also emphasize that $|\exp (i\beta)|=1$ when $\beta$ and $x$ are real, so if $r=\alpha +i\beta$ then $|\exp(xr)|=|\exp(x\alpha)=|e^{-xa_1/2}|=e^{-xa_1/2}$ for real $x.$ Feb 19, 2017 at 23:53
• My edit was to add $=$ signs at the end of one line and the start of the next, in Part 1. Feb 19, 2017 at 23:55
• @user254665 Refrain from editing as such. Equal signs do not belong at the end of the line. Feb 20, 2017 at 0:12
• There were no equal signs between the two lines at all. And you will find it is quite common now to put them at both ends and beginnings of lines Feb 20, 2017 at 0:17

Split the exponentials into a product. For $a_1 > 0$ $e^{-a_1 x/2} \to 0$ as $x \to \infty$.

• Why aren't you taking the limit of the whole solution? Feb 19, 2017 at 17:20
• >$\varphi(x) = c_1e^{(\frac{-a_1}{2}+\frac{\sqrt{a_1^2-4a_2}}{2})x} + c_2e^{(\frac{-a_1}{2}-\frac{\sqrt{a_1^2-4a_2}}{2})x}$ Feb 19, 2017 at 17:20
• Since the characteristic equation has complex roots $\alpha \pm i \beta$ the solution has the form $y = C_1 e^{\alpha x} \sin (\beta x) + C_2 e^{\alpha x} \cos (\beta x)$. Apply the Squeeze Theorem and you'll find each term goes to $0$. Feb 19, 2017 at 17:26

The characteristic polynomial's roots have sum $-a_1<0$ and product $a_2>0$, so are both negative. The solution is therefore exponentially suppressed and vanishes as $x\to\infty$. (Note that if $a_2<0$ we would instead have a positive root, giving a solution that diverges for large $x$, while if $a_2=0$ we have $y_1=A+Be^{-a_1 y}$ for constants $A,\,B$, agreeing with the required result iff $A=0$.)