Question

I desperately need help with this question.

d^2x/dt^2 + dx/dt + Sin x = 0

Find the fixed points

Thanks

Answer 1

Welcome to Open Study @Harrisoncd2468

Answer 2

\[\begin{align*}\frac{d^2x}{dt^2}+\frac{dx}{dt}+\sin x&=0\\\\
\frac{d^2x}{dt^2}+\frac{dx}{dt}&=-\sin x\end{align*}\]
The homogeneous solution:
\[x''+x'=0\]
gives the characteristic equation
\[r^2+r=0~~\iff~~r(r+1)-0~~\iff~~r=0,~r=-1\]
Solution: \(x_h=C_1e^{0x}+C_2e^{-x}=C_1+C_2e^{-x}\)

The nonhomogeneous solution:
Solve by the method of undetermined coefficients. Suppose \(x_p=A\sin t+B\cos t\) is a solution, then
\[\begin{align*}x_p&=A\sin x+B\cos x\\
{x_p}'&=A\cos x-B\sin x\\
{x_p}''&=-A\sin x-B\cos x\end{align*}\]
Substituting into the DE, you have
\[\begin{align*}(-A\sin x-B\cos x)+(A\cos x-B\sin x)&=-\sin x\\
(-A-B)\sin x+(A-B)\cos x&=-\sin x
\end{align*}\]
which gives the system
\[\begin{cases}-A-B=-1\\
A-B=0\end{cases}~~\Rightarrow~~A=B=\frac{1}{2}\]
So, the general solution is
\[\begin{align*}x(t)&=x_h+x_p\\
&=C_1+C_2e^{-x}+\frac{1}{2}\sin x+\frac{1}{2}\cos x
\end{align*}\]

Answer 3

Not too sure about this "fixed point" business. I haven't come across the term before.