1)find two non-zero solutions that aren't multiples of each other 2) sketch the direction field in the yv-plane 3) for each solution, plot both its solution curve in the yv-plane and its y(t) and v(t) graph
1) $\frac{d^{2}y}{dt^{2}}+3\frac{dy}{dt}-10y=0$
$y(t)=e^{st}$
$s^{2}+3s-10=0$
$s=-5, s=2$
$y(t)=k_{1}e^{2t}+k_{2}e^{-5t}$ $k$ is a constant.
One solution is: $y(t)=3e^{2t}+2e^{-5t}$
another is: $y(t)=5e^{2t}+4e^{-5t}$
2)
The direction field is a saddle at the origin.
3) Here is where i am confused. Would the solution curve in the yv plane be graphs of $\frac{dv}{dt}=-3v+10y$ and $\frac{dy}{dt}=v$
or the graphs of the two solutions from part 1) on the same axis.
Part 1: In order to not confuse variable names, rewrite the system as:
$$Y''(t) + 3 Y'(t) -10 Y(t) = 0$$
Let $Y = y$, so
$$y' = v\\v' = 10 y - 3v$$
The eigenvalues are $\lambda_1 = -5, \lambda_2 = 2$, with eigenvectors $v_1 = (1,-5), v_2 = (1,2)$.
This gives,
$$Y(t) = \begin{bmatrix} y(t) \\ v(t) \\ \end{bmatrix} = c_1 e^{-5 t}\begin{bmatrix} 1 \\ -5 \\ \end{bmatrix} + c_2e^{2 t}\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$$
We will choose two different initial conditions to get two non-zero solutions.
For $IC1$, choose $y(0) = 0, v(0) = 1$, leading to
$$Y(t) = \begin{bmatrix} y(t) \\ v(t) \\ \end{bmatrix} = \begin{bmatrix} ~~~\dfrac{2}{7}~~ e^{-5t}~~ + ~\dfrac{5}{7}~e^{2t} \\ -\dfrac{1}{7} e^{-5t} + \dfrac{1}{7}e^{2t} \\ \end{bmatrix}$$
For $IC2$, choose $y(0) = 1, v(0) = 0$, leading to
$$Y(t) = \begin{bmatrix} y(t) \\ v(t) \\ \end{bmatrix} = \begin{bmatrix} ~~~\dfrac{2}{7}~~ e^{-5t}~~ + ~\dfrac{5}{7}~e^{2t} \\ -\dfrac{10}{7} e^{-5t} + \dfrac{10}{7}e^{2t} \\ \end{bmatrix}$$
Part 2: We have a critical point at $(v, y) = (0, 0)$ and it is a saddle point from the eigenvalues. Using several hand calculations (see Example 1) for the slope field direction gives
Part 3:
If we parametrically plot $y(t)$ versus $v(t)$ for $IC1$ (red) and $IC2$ (purple) we have:
Lastly, we plot solutions for $IC1$ and $IC2$ onto the phase portrait:
Do you notice why the phase portrait is so useful from these two solutions shown on top of it?