17.1.08 y''+100y=0 Find the general solution of the given equation?

[karush]

$\tiny{17.1.08}$
$\textrm{ Find the general solution of the given equation?}$
\begin{align*}\displaystyle
y''+100y&=0
\end{align*}
$\textit{The auxiliary equation is:}$
\begin{align*}\displaystyle
x^2+100x&=0\\
x(x+100)&=0\\
x&=-100
\end{align*}
$\textit{Answer by EMH}$
\begin{align*}\displaystyle
y&=c_1 \cos(10x)+c_2 \sin(10x)
\end{align*}
ok tried looking at some examples again but not?

this doesn't have imaginary roots

 

[MarkFL]

The characteristic equation is:

$$r^2+100=r^2+10^2=0$$

Thus:

$$r=\pm10i$$ :D

 

[karush]

so
$$y=c_1 \cos(10 x) + c_2 \sin(10 x))$$

 

[MarkFL]

so
$$y=c_1 \cos(10 x) + c_2 \sin(10 x))$$

Yes, what if the ODE had been:

$$y''-2y'+11y=0$$

Here you find the roots of the characteristic equation to be:

$$r=1\pm\sqrt{10}i$$

What would the solution then be?

 

[HallsofIvy]

You are asking Karush whether he can answer that, not because you can't do it, right?

 

[MarkFL]

You are asking Karush whether he can answer that, not because you can't do it, right?

Yes, I wanted to see if the OP knew how to handle characteristic roots that have both real and imaginary components (complex roots in general). :D

 

[karush]

The characteristic equation is:

$$r^2+100=r^2+10^2=0$$

Thus:

$$r=\pm10i$$ :D

isn't it??

$r^2+10^2 r =0$

 

[MarkFL]

isn't it??

$r^2+10^2 r =0$

No, when creating the characteristic equation the order of differentiation translates to the exponent on the characteristic variable, i.e.:

$$y^{(n)}(x)\implies r^n$$

Therefore $y(x)\implies r^0=1$.

 

[HallsofIvy]

You can get the "characteristic equation" for a linear differential equation with constant coefficients by looking for a solution of the form y(x)= e^{rx}.

Then y'(x)= re^{rx} and y''(x)= r^2e^{rx}. So y''+ 100y= r^2e^{rx}+ 100e^{rx}= (r^2+ 100)e^{rx}= 0. Because e^{rx} is never 0, we must have r^2+ 100= 0 which has roots \pm 10i.

You only have an "r" in the characteristic equation when there is a y' in the differential equation.