SM212, PROF JOYNER, 9-14-98
TEST 1

PROBLEM 1 (a) y'=x-y, y(0) = 1.
(B) For y as in part (a), compute y(1) exactly.

Solution: y' + y = x
p=1    q=x

y=[int( eint 1 dx x dx) + c]/eint(1 dx)   =[int( ex x dx) + c]/ex   = (xex   - ex + c)/ex .

1=0-1+Ce0           C=2             y=1-1+2e-1              y(1)= 2/l             y= x - 1 + 2e-x

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

Solution:

x                  y                  1/6[x-y + (y+1/3(x-y)))]
__________________________________

0                 1                            -2/9
1/3            7/9                       -11/162
2/3        115/162                 127/2916
1          2197/2916
y(1)= .7534

PROBLEM 3 Wiley E. Coyote is flying through the air drinking a can of cold Cola at 40 F with his ACME jet pack trying to catch the Road Runner then his fuel runs out.  He falls in a brine solution without spilling a drop.  The air temperature and the temperature of the solution was a constant 70 F.  After 10 minutes the cola warmed up to 60 F.  How long did it take to cool to 50 F?

Solution: T(0)=40       T(t) = Troom = Cekt
Troom=70              40 = 70 + Ce0
T(10)=60              -30 = C
T(?)= 50                60 = 70 - 30e10k
ln(1/3)= 10k
1/10 ln(1/3) = k
50= 70 -30e(1/10) t ln(1/3)
2/3 =e (1/10) t ln(1/3)
ln(2/3)/(1/10 ln (1/3) )   =t             t= 3.69 minutes

PROBLEM 4 Solve yy'= x2 + 1, y(0) = -1, explicitly (for y as a function of x).

Solution:

dy *y = x2+1            ydy=(x2+1)dx
dx                               (x2+1)/((1/3)x3 +x)dx     =       (-y)/(-1/2y2)dy               0=0 exact

y(0) = -1                           -1/2y2 + 1/3x3 + x = c
-1 =  (2c)(-1/2)               -y2= 2(-1/3x3 - x + c)
1=2c                                  y = -square root( [2(1/3x3+x+c)])
c=1/2                                 y= - square root([2/3x3 + 2x +1])

PROBLEM 5 Curly, Moe and Larry are incompetent butlers for Daffy Duck, who orders them to prepare his bath.  The bathtub holds 100 gallons when filled.  Curly begins pouring in a soap solution at a rate of 1 gallon/minute.  The soap concentration is 1/4 lb/gallon in his soap water solution.  Larry put the plug in the drain wrong so that water drains out at ½ gal/min.  Originally, Moe only filled the bath half full (of pure water).  Find the amount (in lbs) of soap in the bathtub when it is full.

Solution:
Cin = 1/4 lb/gal
Sin = 1 gal/min
Sout = ½ gal/min            P= 1/100+1     A' = RATEin - RATEout
A(0) = 0 lbs                   Q= 1/4             A' = CinSin - CoutSout
A(100) = ?                                             A' = 1/4(1) - 1/2[(A(t)/50+(1/2)t)]
A' +(1/100+t)(A(t)) = 1/4
Cout= A(t)/W(t)                                     A = 1/8(100)2+25(100)
200
A = 18.75 lbs
A(t) = (int(eint( 1/(100+t) dt)*1/4 dt) + c)/(  eint(1/(100+t) dt) )

A= (int.(eln(t+100)* 1/4 dt+c)/e ln(t + 100)

A= [1/4 int( t + 100 dt) + c] /( t + 100 )

= [1/4(1/2 t2 + 100t) + c]/(   t + 100 )

= [1/8t2 + 25t + c]/(   t + 100 )

0 =  c/100   c=0

BONUS Solve y' = (y4 - 1), y(0) = 0

Solution:

dy/dx = y4 - 1                   1/(y4-1)dy   = dx

dx                            y4 - 1dy                                    0=0 exact

x               ln((y-1)/(y+1)) - arctan(y)
4                     2

x+ ln(y-1/y+1)  - arctan(4)   = c
4                 2

c = 0

x + ln(y-1/y+1)    - arctan(y) = 0
4                       2

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