__SM212, Professor Joyner__

**TEST 1, 5 FEB 1997**

**PROBLEM 1** (a) Solve *y' = 2xy + x, y(0) = 1*.

(b) For *y* as in part (a), compute *y(1)* exactly.

**
PROBLEM 2
**
Using *h* = 1/3 in Euler's method, approximate *y*(1),
where *y' = 2xy +x*, *y*(0) = 1

Answer:

x | y | 1/3*(2xy+x) |

0 | 1 | 0 |

1/3 | 1 | 1/3 |

2/3 | 4/3 | 22/27 |

1 | 58/27 |

y(1) is approx 58/27

**
PROBLEM 3
**
Using *h* = ½ in improved Euler's method, approximate
*y*(1), where *y' = 2xy + x*, *y*(0) = 1.

Answer:

x | y | h*(f+f^-)/2=(2xy+x+2(x+1/2)(y+(1/2)(2xy+x))+x+1/2)/4 |

0 | 1 | 3/8 |

(1/2) | 11/8 | 15/8 |

1 | 13/4 |

y(1) is approx 13/4=3.25.

**
PROBLEM 4
**
Solve *y' = sec(x) ^{2}( 1 + y^{2})*,

**
PROBLEM 5
**
A can of coke at 40 deg F is placed in a room of constant
temperature 70 deg F. After 10 minutes it warms up to 60 deg F. How long
did it take to cool to 50 deg F?

**
PROBLEM 6
**
Using direction fields and at least 3 isoclines, sketch the
solution to *y' = 2xy + x*,

*y*(0) = 1. Label the isoclines with their slope.

Hint:

Isocline : {(x,y) | 2xy + x = m}

We take m to equal: -1, 0 , 1. These give: m=0: {x=0 or y=-1/2}. m=1: {y=-1/2+1/(2x)} (shifted hyperbola with asymptotes the m=0 isocline). m=-1: {y=-1/2-1/(2x)} (shifted hyperbola with asymptotes the m=0 isocline).

To complete the answer, graph the above equations on the same coordinate system, labeling each isocline with its slope. (The slope is the m value for each equation). Then, starting at (0,1), (the given condition, y(0) = 1) graph the solution using your previously drawn slope fields.

Looks like an upward parabola based at (0,1).Good Luck!

Copyright 1997. Written by Midn 3/C Pete L. Flores on 21 April 1997. Last modified 9-14-2004 by wdj.