Question

What is the general solution for r^6+16r^2=0?

Answer 1

$$\Large{
r^6+16r^2=0?
\\r^2( r^4 + 16) = 0
\\ \Rightarrow r^2 = 0 ~,~ r^4 + 16 = 0
}$$

Answer 2

r^4 = -16 im stuck after that

Answer 3

that does not have real solutions,

Answer 4

are you looking for the imaginary solutions as well

Answer 5

yeah

Answer 6

\[
r^4=-16\\
r=2\sqrt[4]{-1}
\]
So find the quartic root of -1.

Answer 7

+/- 2i?

Answer 8

right

Answer 9

I need the answer in the form y=c1+c2... but I only know r1 and r2=0 and r4,r5,r6 are missing.

Answer 10

this is a differential equation, can you state the original problem

Answer 11

Find the general solution to y^6+16y''=0

Answer 12

I don't think the answer are -2i and 2i and (-2i)^4=(2i)^4=16.

Answer 13

yeah thats what i was thinking then id still have r6 missing

Answer 14

r^4 = -16
r^2 = ± sqrt(-16)
r^2 = ± 4i
r = ± √ (± 4i)

Answer 15

should i use -1= e^i(pi+2npi) for that

Answer 16

\[
-1=\cos(\pi)+i\sin(\pi)\\
\sqrt[4]{-1}=\cos\left(\frac{\pi}{4}+\frac{n\pi}{2}\right)+i\sin\left(\frac{\pi}{4}+\frac{n\pi}{2}\right),\,n=0,1,2,3
\]

Answer 17

gotcha thanx!

Answer 18

r^4 = -16
r^2 = ± sqrt(-16)
r^2 = ± 4i
r = ± √ (± 4i)

r= √( 4i), √(-4i), -√(4i), -√(-4i)

Answer 19

r^4 = -16
r^2 = ± sqrt(-16)
r^2 = ± 4i
r = ± √ (± 4i)
r= 2√i, 2√-i, -2√i, -2√i

Answer 20

@perl would doing that thomas' way give me the same answer?

Answer 21

sure :)

Answer 22

got it thank you guys!

Answer 23

√i = √(2)/2 + i √(2)/2

Answer 24

Demoivre's theorem is an good approach to finding the nth root of a number

Answer 25

I have no idea what I am doing but could we try this?
\[
y^{(6)}+16y''=0\\
y^{(5)}+16y'=a\\
y^{(4)}+16y=ax+b\\
\lambda^4+16=0\\
\lambda=\frac{\sqrt{2}}{{2}}+i\frac{\sqrt{2}}{{2}},\,-\frac{\sqrt{2}}{{2}}+i\frac{\sqrt{2}}{{2}},-\frac{\sqrt{2}}{{2}}-i\frac{\sqrt{2}}{{2}},\frac{\sqrt{2}}{{2}}-i\frac{\sqrt{2}}{{2}}\\
y=c_1e^{\left(\frac{\sqrt{2}}{{2}}+i\frac{\sqrt{2}}{{2}}\right)x}+c_2e^{\left(-\frac{\sqrt{2}}{{2}}+i\frac{\sqrt{2}}{{2}}\right)x}+c_3e^{\left(-\frac{\sqrt{2}}{{2}}-i\frac{\sqrt{2}}{{2}}\right)x}+c_4e^{\left(\frac{\sqrt{2}}{{2}}-i\frac{\sqrt{2}}{{2}}\right)x}+ax+b
\]

Answer 26

Just in case you can't see this
\[
\begin{align*}
y&=c_1e^{\left(\frac{\sqrt{2}}{{2}}+i\frac{\sqrt{2}}{{2}}\right)x}+c_2e^{\left(-\frac{\sqrt{2}}{{2}}+i\frac{\sqrt{2}}{{2}}\right)x}\\
&\quad +c_3e^{\left(-\frac{\sqrt{2}}{{2}}-i\frac{\sqrt{2}}{{2}}\right)x}+c_4e^{\left(\frac{\sqrt{2}}{{2}}-i\frac{\sqrt{2}}{{2}}\right)x}\\
&\quad +ax+b
\end{align*}
\]

Answer 27

thanx again @thomas5267 i appreciate it!