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.

Answer: Integerating factors gives y=(-e^(-x^2)/2+C)/e^(-x^2)= -1/2+Ce^(x^2). Since y(0)=1, we must have C=3/2, so y=-1/2+(3/2)e^(x^2). y(1)=-1/2+(3/2)e=3.577...

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

 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.

 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 + y2), y(1) = pi/4.

Answer: arctan(y)=tan(x)+C. Use y=Pi/4 and x=1 to solve for C: get C=(4-Pi)/4.

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?

Answer: Troom=70, T(0)=40,T(10)=60, so T(t)=70-30*(1/3)^(t/10), so t=ln(2/3)*10/ln(1/3)=3.69...

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.