Hi guys! I think I posted a question similar to this before. I think i've made some progress in solving questions of this type, but i'm still getting stuck on bits of it.

The method for this is long so i'll just put the answer.Quote:

Find the general solution to $\displaystyle \ddot{x}-2\dot{x}-3x=10 \sin t$.

This comes out to be $\displaystyle x(t)=Ae^{3t}+Be^{-t}-2 \sin t + \cos t$ where $\displaystyle A,B$ are constants.

$\displaystyle \dot{x}=y$Quote:

Write this equation in the three variables $\displaystyle x,y=\dot{x}$ and $\displaystyle \dot{s}=1$ so that the equations are autonomous and first order.

$\displaystyle \dot{y}=10 \sin s +3x+2y$

$\displaystyle \dot{s}=1$

We know that $\displaystyle \frac{ds}{dt}=1$. This gives $\displaystyle \frac{dx}{dt}=\frac{ds}{dt}. \frac{dx}{ds}=\frac{dx}{ds}$.

Applying a similar method gives:

$\displaystyle \frac{dy}{dt}=\frac{ds}{dt}.\frac{dy}{ds}=\frac{dy }{ds}$

We also have:

$\displaystyle

\frac{d^2x}{dt^2} = \frac{d}{dt} \left(\frac{dx}{dt}\right) = \frac{d}{ds} \left( \frac{dx}{ds}\right) \cdot \frac{ds}{dt} = \frac{d^2x}{ds^2} = \frac{dy}{ds}

$ since $\displaystyle y=\frac{dx}{ds}$.

This gives my 3 equations as:

$\displaystyle \frac{dt}{ds}=1$

$\displaystyle \frac{dx}{ds}=y$

$\displaystyle \frac{dy}{ds}-2y-3x=10 \sin t$

(Kudos to Danny for this!)

$\displaystyle x(p) \in \mathbb{R}^3$Quote:

Write down the corresponding solution $\displaystyle (x(t),y(t),s(t))$.

$\displaystyle x(p)=(\dot{x}, \dot{y},\dot{s})=(y,10 \sin s+3x+2y,1)$

But we know the general solution.

$\displaystyle (x(t),y(t),s(t))=(Ae^{3t}-2 \sin t+ \cos t, 3Ae^{3t}-2 \cos t - \sin t, t+C)$

So we're going from $\displaystyle s=0$ to $\displaystyle s= 2 \pi$Quote:

The vector field in $\displaystyle \mathbb{R}^3$ is now periodic in $\displaystyle s$ with period $\displaystyle 2 \pi$.

Write down the Poincare map in the plane $\displaystyle s= 2 \pi$ for a set of initial conditions $\displaystyle (x_0,y_0)$ lying in the plane $\displaystyle s=0$.

$\displaystyle (x(0),y(0),0)=(x_0,y_0,0)$

This gives:

$\displaystyle A+1=x_0$

$\displaystyle 3A-2=y_0$

$\displaystyle C=0$

$\displaystyle \Rightarrow 4A-1=x_0+y_0$

$\displaystyle \Rightarrow A=\frac{1}{4}(x_0+y_0+1)$

So this gives:

$\displaystyle (x(t),y(t),s(t))=$$\displaystyle \left( \frac{1}{4}(x_0+y_0+1)e^{3t}-2 \sin t+ \cos t, \frac{3}{4}(x_0+y_0+1)e^{3t}-2 \cos t- \sin t, t \right)$

This is where I get stuck. I only have one example in my notes that does not have an $\displaystyle e^{3t}$ in it. As for the part on eigenvalues, I have no idea how to even get a matrix from this!!Quote:

Find it's fixed point and discuss it's stability in terms of eigenvalues.

I'd appreciate some help on this, there are still more bits of the question that I have to try and do!!