Show that no values of $b$ can make the equation $b\sin(bx)-2\cos(bx) = 0$ true. I am trying to prove that $()=\cos()$ is a solution to the DE $'''+2''+'+2=0$, and by substituting $f$
and its derivatives, I have simplified the equation to $$(^2−1)(\sin()−2\cos())=0.$$ It is easy to show $^2−1=0$ 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?
 A: 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?
A: $$(^2−1)(\sin()−2\cos())=0.$$
$$(\sin()−2\cos())=0.$$
$$\implies \tan (bx)=\dfrac  2b \,\, (\forall \,\, x) $$
Can't be true since $x$ is variable and $\dfrac 2b $ is a constant. It can only be true for a certain value of $x$.
A: You have to justify it.
When $x=0$, we have $-2=0$ regardless of the value of $b$, hence no such $b$ exists.
A: $\cos bx$ 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 $y = c_1e^{-2x} + c_2 \cos x + c_3 \sin x$.
