9 Ordinary Differential Equations I

In this class,

  1. For the ODE

    \begin{displaymath}
y' = x(1-y)
\end{displaymath}

    sketch a dense direction field as a fully covering set of tiny line segments. Or use Mathlab or similar to draw the segments. Based on that, draw various solution curves. Discuss maxima and minima, symmetry, asymptotes, and inflection points of the solutions. Do NOT solve the equation algebraically. Use only the direction field to derive the solution properties. No cheating!

  2. Solve

    \begin{displaymath}
2 y \frac{{\rm d} y}{{\rm d} x} = e^x e^{-y^2}
\qquad
y(4)=-2
\end{displaymath}

    Now solve the same ODE, but with initial condition that $y=0$ at $x=1$. Accurately draw these solutions.

  3. Solve, using the class procedure (variation of parameter),

    \begin{displaymath}
\sin(2x) \frac{{\rm d} y}{{\rm d} x} = 2 \sin(x) - 2 y \sin^2(x)
\end{displaymath}

    Draw a few representative solution curves.

  4. Solve, using the class procedure,

    \begin{displaymath}
\frac{{\rm d} y}{{\rm d} x} = \frac{y}{x+\sqrt{xy}}
\end{displaymath}

    Draw a few representative solution curves. (Hint: solve for $x$ as a function of $y$.) Also find the solution curve that satisfies the initial condition $y(0)=1$ and draw it in your graph in a different color.

  5. Solve, using the class procedure,

    \begin{displaymath}
y' + \frac{2}{x} y = - x^9 y^5
\end{displaymath}

    Draw a few representative solution curves. Also find the solution curve that satisfies the initial condition $y(-1)=2$ and draw it in your graph in a different color. Where is its vertical asymptote?

  6. Solve using class procedures

    \begin{displaymath}
y'' + 2 y' - 3 y = 0 \qquad y(0) = 6 \qquad y'(0) = -2
\end{displaymath}