9 Ordinary Differential Equations I

In this class,

• Questions must be answered in order asked.
• Solutions must be neat.
• Solutions must be exact and fully simplified.
• You must use the given symbols.
• You must show all reasoning.
• Copying is never allowed, even when working together.

1. For the ODE
   
   $$y' = x(1-y)$$
   
   
   sketch a dense direction field as a fully covering set of tiny line segments. A few hundred line segments will do, if you choose their locations well. Or use the Matlab quiver(u,v,scale) function or similar to draw the segments. (If $dy/dx=\tan\alpha$, you could take $u=\cos\alpha$ and $v=\sin\alpha$) Based on that, draw various solution curves accurately. 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. Cheating reduces credit!

2. Solve
   
   $$2 y \frac{{\rm d} y}{{\rm d} x} = e^x e^{-y^2} \qquad y(4)=-2$$
   
   
   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),
   
   $$\sin(2x) \frac{{\rm d} y}{{\rm d} x} = 2 \sin(x) - 2 y \sin^2(x)$$
   
   
   Draw a few representative solution curves.

4. Solve, using the class procedure,
   
   $$\frac{{\rm d} y}{{\rm d} x} = \frac{y}{x+\sqrt{xy}}$$
   
   
   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,
   
   $$y' + \frac{2}{x} y = - x^9 y^5$$
   
   
   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
   
   $$y'' + 2 y' - 3 y = 0 \qquad y(0) = 6 \qquad y'(0) = -2$$
   
   

7. Solve using class procedures
   
   $$2 \stackrel{...}{x} - 10 \stackrel{..}{x} + 16 \stackrel{.}... ... x(0) = \stackrel{.}{x}(0) = 0 \qquad \stackrel{..}{x}(0) = 2$$