Question

x^3y'''+3x^2y''+xy'=y=0.
Can someone please show me how to find all real solutions of this? I changed it by making y=x^m and came up with m^3+1=0 which gives the repeated roots -1, -1, -1....then it gets ugly and I have no clue what I've done wrong.

Answer 1

\[\begin{cases}
y=x^m\\
y'=mx^{m-1}\\
y''=m(m-1)x^{m-2}\\
y'''=m(m-1)(m-2)x^{m-3}
\end{cases}\]
So\[x^3y'''+3x^2y''+xy'+y=0\]
becomes
\[\begin{align*}x^3\left(m(m-1)(m-2)x^{m-3}\right)+3x^2\left(m(m-1)x^{m-2}\right)+x\left(mx^{m-1}\right)+x^m&=0\\
x^m\bigg(m(m-1)(m-2)+3m(m-1)+m+1\bigg)&=0\\
m^3+1&=0
\end{align*}\]
which has only one real root. The other two are complex.

Answer 2

Let's forget about finding the actual complex roots for a moment, ad focus on how to deal with them.

Suppose we have a complex root \(m=a+bi\), which means you have a solution of \(y=x^m=x^{a+bi}\). With some algebraic manipulation you can find a more useful expression:
\[\large x^m=x^{a+bi}~~\iff~~x^m=e^{\ln x^{a+bi}}\]
Some log and exponent properties as well as Euler's identity yield
\[\large\begin{align*}
x^m&=e^{(a+bi)\ln x}\\
&=e^{a\ln x}e^{bi\ln x}\\
&=e^{\ln x^a}\bigg(\cos (b\ln x)+i\sin(b\ln x)\bigg)\\
&=x^a\bigg(\cos (b\ln x)+i\sin(b\ln x)\bigg)\end{align*}\]
All this means your general solution will look like
\[y=x^m=C_1x^a\cos(b\ln x)+C_2x^a\sin(b\ln x)\]

Answer 3

ok so do I still need to find the complex roots or do I only need to worry about the one real root.

Answer 4

You can't ignore the complex roots! Unless your directions ask for the real-root solution only, you must provide *all* the solutions to the differential equation.

Now getting back to root-finding... Recall that -1 is a complex number. Let's write it in polar/trigonometric form:
\[-1=-1+0i\]
which gives
\[\begin{cases}r\cos\theta=-1\\
r\sin\theta=0\end{cases}~~\implies~~r=1,~\theta=\pi\]
With the help of DeMoivre's theorem, we can verify that all the \(n\)th roots of a complex number \(\bigg(\)i.e. all complex solutions to \(m^n=a+bi=re^{i\theta}\bigg)\) have the form,
\[\Large m=r^{1/n}e^{i(\theta+2k\pi)/n}\]
where \(k=0,1,...,n-1\).

In this case, \(\large m^3=a+bi=-1+0i=e^{i\pi}\), so \(n=3\), and \(k=0,1,2\).

We thus have the following cubic roots:
\[\Large\begin{align*}m_1&=(1)^{1/3}e^{i(\pi+2\pi)/3}=e^{3i\pi/3}=e^{i\pi}\\\\&=-1\\\\
m_2&=(1)^{1/3}e^{i(\pi+4\pi)/3}=e^{5i\pi/3}=\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\\\\
&=\frac{1}{2}-i\frac{\sqrt3}{2}\\\\
m_3&=(1)^{1/3}e^{i(\pi+6\pi)/3}=e^{7i\pi/3}=\cos\frac{7\pi}{3}+i\sin\frac{7\pi}{3}\\\\
&=\frac{1}{2}+i\frac{\sqrt3}{2}
\end{align*}\]

Answer 5

Sorry for the obnoxious text size, I didn't think it'd show :P

Answer 6

but it does say to find all real solutions. I assumed that meant no complex ones!

Answer 7

Alright, in that case you can ignore all that extra work. It's still good practice to find the complex ones, but directions are directions.

Answer 8

At any rate, there are no repeated roots!

Answer 9

Do you think I should still include them? Will I need to do the Wronskian and all that/

Answer 10

so y1= 1/x y2= (1/2)(cos(\[\sqrt{3}/2\ln x\]
and then y3 is the same as y2 only it's sin? I'm sorry I'm not good with this equation tool

Answer 11

If you're only asked to find the real-root solution, you're done when you find \(y_1\). There's no need to check the Wronskian if you only have one solution.

On the other hand, if you want to include \(y_2\) and \(y_3\) (which are not quite correct, by the way), it should be clear that they're linearly independent (what some might call the eyeball test), but you're free to check the Wronskian just to be sure.

The actual solutions for \(y_2\) and \(y_3\) are
\[y_2=x^{1/2}\cos\left(\frac{\sqrt3}{2}\ln x\right)\]
and \(y_3\) has cos switched with sin.

Answer 12

OH yeah!!! I had that written down I'm sorry I just wrote it on here wrong!

Answer 13

Thanks so much!

Answer 14

You're welcome!