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I am trying to prove that $𝑓(𝑥)=\cos(𝑏𝑥)$ is a solution to the DE $𝑦'''+2𝑦''+𝑦'+2𝑦=0$, and by substituting $f$ and its derivatives, I have simplified the equation to $$(𝑏^2−1)(𝑏\sin(𝑏𝑥)−2\cos(𝑏𝑥))=0.$$ It is easy to show $𝑏^2−1=0$ has appropriate values for $b$ , but I am not so sure about $b\sin(bx)−2\cos(bx)=0$.

If I have the equation $b\sin(bx)-2\cos(bx) = 0$, how do I prove that no values of $b \in \mathbb{R}$, can make the equation true? It it enough to say that $2\cot(bx) \neq b$ for all $x \in \mathbb{R}$, hence $b$ does not exists?

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  • $\begingroup$ This statement is false. The only value of $b$ where this happens is $b = 0$. $\endgroup$
    – Toby Mak
    Apr 7, 2021 at 2:12
  • $\begingroup$ This sounds like an XY problem: what is your real question? $\endgroup$
    – Toby Mak
    Apr 7, 2021 at 2:12
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    $\begingroup$ @TobyMak I am trying to prove that $f(x) = \cos(bx)$ is a solution to the DE $y'''+ 2y'' + y' +2y = 0$, and by substituting $f$ and its derivatives, I have simplified the equation to $(b^2-1)(b\sin(bx)-2\cos(bx))=0$. It is easy to show $b^2 -1 = 0$ has appropriate values for $b$, but I am not so sure about $b\sin(bx) - 2\cos(bx) = 0$. $\endgroup$
    – sinclair
    Apr 7, 2021 at 2:18
  • $\begingroup$ That sounds much better, so thank you for adding this information. If you could copy that into the question body, I'll remove the downvote. $\endgroup$
    – Toby Mak
    Apr 7, 2021 at 2:19

4 Answers 4

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HINT

Here it is an alternative way to approach it for the sake of curiosity.

You can actually solve the proposed ODE:

\begin{align*} y''' + 2y'' + y' + 2y = 0 & \Longleftrightarrow (y''' + 2y'') + (y' + 2y) = 0\\\\ & \Longleftrightarrow (y' + 2y)'' + (y' + 2y) = 0\\\\ & \Longleftrightarrow u'' + u = 0\\\\ & \Longleftrightarrow u(x) = c_{1}\cos(x) + c_{2}\sin(x)\\\\ & \Longleftrightarrow y' + 2y = c_{1}\cos(x) + c_{2}\sin(x)\\\\ & \Longleftrightarrow (\exp(2x)y)' = \exp(2x)(c_{1}\cos(x) + c_{2}\sin(x)) \end{align*}

Can you take it from here?

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$$(𝑏^2−1)(𝑏\sin(𝑏𝑥)−2\cos(𝑏𝑥))=0.$$ $$(𝑏\sin(𝑏𝑥)−2\cos(𝑏𝑥))=0.$$ $$\implies \tan (bx)=\dfrac 2b \,\, (\forall \,\, x) $$ Can't be true since $x$ is variable and $\dfrac 2b $ is a constant. It can only be true for a certain value of $x$.

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You have to justify it.

When $x=0$, we have $-2=0$ regardless of the value of $b$, hence no such $b$ exists.

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$\cos bx$ is not a solution, as after differentiating once, you will get $-b \sin bx$, and the only value of $b$ that will cancel the $\sin$ terms is $0$. You might then try $A \cos bx + B \cos bx$, but there is a much simpler way.

Find the characteristic polynomial by substituting in $y = e^{ax}$:

$$r^3 + 2r^2 + r + 2 = 0 \Rightarrow r^2(r+2) + (r+2) = 0$$ $$\Rightarrow (r^2+1)(r+2) = 0, r = -2, ±i$$

and since $\sin x$ and $\cos x$ can be expressed as linear combinations of $e^{ix}, e^{-ix}$, the particular solution is $y = c_1e^{-2x} + c_2 \cos x + c_3 \sin x$.

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