Do I have to use Euler's formula when solving this Euler-Cauchy equation? I'm currently studying ODE's (Euler-Cauchy equations to be exact) using Advanced Engineering Mathematics (Kreyszig, 2006) and came across an exercise problem and its solution that I'm a bit confused about. I believe that I solved it correctly, but my answer differs from the author's and I'm curious if it's actually incorrect or if it's just a matter of form, and what the intuition behind the different solution is. The problem is:

Solve the following Euler-Cauchy equation:
$$ x^2y''+0.6xy'+16.04y=0 $$

My Solution:
Setting $y = x^m$ as a general solution gives us the characteristic equation of the Euler-Cauchy formula:
$$m^2 -0.4m + 16.04 = 0$$
The roots of this equation are:
$$
\begin{align}
m_1 & = 0.2 + 4i \\
m_2 & = 0.2 - 4i
\end{align}
$$
Since the characteristic equation has conjugate complex roots, we can assume a general solution form of:
$$y = e^{-ax / 2} \left( A\cos{(\omega x)} + B\sin{(\omega x)} \right)$$
where $a = -0.4$, $b = 16.04$, and $\omega^2 = b - \frac{1}{4}a^2$ and so $\omega = 4$.
The final general solution and hence the answer is:
$$y = e^{0.2x}\left( A\cos{(4x)} + B\sin{(4x)} \right)$$
Textbook Solution:
Obtaining the roots of the characteristic equation are the same, but where this solution differs is when the author uses a supposed "trick" of setting $x = e^{\ln{(x)}}$. According to the author this gives:
$$
\begin{align}
x^{m_1} & = x^{0.2 + 4i} = x^{0.2}\left(e^{\ln{(x)}}\right)^{4i} = x^{0.2}e^{(4\ln{(x)})i} \\
x^{m_2} & = x^{0.2 - 4i} = x^{0.2}\left(e^{\ln{(x)}}\right)^{-4i} = x^{0.2}e^{-(4\ln{(x)})i}
\end{align}
$$
Next, the author applies Euler's formula ($e^{it} = \cos{(t)} + i\sin{(t)}$) and by setting $t = 4\ln{(x)}$ you get the solutions:
$$
\begin{align}
x^{m_1} & = x^{0.2} \left( \cos{(4\ln{(x)})} + i\sin{(4\ln{(x)})} \right) \\
x^{m_2} & = x^{0.2} \left( \cos{(4\ln{(x)})} - i\sin{(4\ln{(x)})} \right)
\end{align}
$$
The next part is where my confusion starts: The author states that "we add these two formulas, so that the sine drops out, and divide the result by 2. Then we subtract the second formula from the first, so that the cosine drops out, and divide the results by $2i$." This gives the two formulas: $x^{0.2} \cos{(4\ln{(x)})}$ and $x^{0.2}\sin{(4\ln{(x)})}$.
Due to the superposition principle of linear homogeneous ODE's, we can obtain the final solution:
$$
y = x^{0.2} \left( A\cos{(4\ln{(x)})} + B\sin{(4\ln{(x)})} \right)
$$
As you can see, the two solutions seem similar yet are different. My questions can be summarized as follows:

*

*Is my solution wrong? When I graph the two solutions with the same constant coefficients, they form different graphs. I'll post the two graphs below. The red is my solution and the blue is the author's solution.


*What is the intuition behind using Euler's formula in this setting? I don't see a particular reason for doing this, and the textbook lacks explanation.


*What does the author mean in the "subtracting adding" part? For example, when the author states that we "add these two formulas, so that the sine drops out, and divide the result by 2," wouldn't this mean that we have $\dfrac{x^{m_1} + x^{m_2}}{2}$ and not $x_{m_1}$, thus resulting in different solutions?
Any tips or feedback are appreciated. Thank you.

 A: You need to be careful with back-substituting exactly in the same way you constructed the forward substitution. You tried to find solutions of the form $y=x^m$. Thus after having found $m=0.2\pm 4i$ you need to insert this exactly the same way, $$y=x^{0.2\pm 4i}.$$
For a real problem you want real solutions. Fortunately, real and imaginary part of the basis solutions are also linear combinations of these basis solutions. This is not that mysterious, for a real equation (that makes sense also over the complex numbers) that has a complex solution $z$, also its complex conjugate $\bar z$ will be a solution. As the equation is also linear, $Re(z)=(z+\bar z)/2$ and $Im(z)=(z-\bar z)/(2i)$ will also be solutions.
It remains to determine what 
$$
\frac{x^{0.2+4i}+x^{0.2-4i}}2=\sqrt[5]x\frac{x^{4i}+x^{-4i}}2
$$
actually mean. Here then comes the transformation into and exponential to the Euler number and the Euler formula into play.

Your original approach also works, but then you have at the start to substitute $x=e^t$, $u(t)=y(e^t)$, $u'(t)=y'(e^t)e^t$, $u''(t)=y''(e^t)e^{2t}+y'(e^t)e^t$ which implies
$$
u''(t)-0.4u'(t)+16.04u(t)=0,
$$
giving the same characteristic polynomial and the formula in the usual exponentials and trigonometric functions, but in variable $t=\log x$, not $x$ itself.
