How to find solutions and to graph this second order DE?

2,078

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

enter image description here

Part 3:

If we parametrically plot $y(t)$ versus $v(t)$ for $IC1$ (red) and $IC2$ (purple) we have:

enter image description here

enter image description here

Lastly, we plot solutions for $IC1$ and $IC2$ onto the phase portrait:

enter image description here

Do you notice why the phase portrait is so useful from these two solutions shown on top of it?

Share:
2,078

Related videos on Youtube

Big Mike
Author by

Big Mike

Updated on December 15, 2020

Comments

  • Big Mike
    Big Mike almost 3 years

    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.

    • Big Mike
      Big Mike almost 8 years
      @evgeny can you please help!