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( e^{int 1 dx} x dx) + c]/e^{int(1 dx)}   =[int( e^x x dx) + c]/e^x   = (xe^x   - e^x + c)/e^x .
              
 

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 = Ce^kt
     Troom=70              40 = 70 + Ce⁰
     T(10)=60              -30 = C
     T(?)= 50                60 = 70 - 30e^10k
                                    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' = RATE_in - RATE_out
A(0) = 0 lbs                   Q= 1/4             A' = C_inS_in - C_outS_out
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(e^{int( 1/(100+t) dt)}*1/4 dt) + c)/(  e^{int(1/(100+t) dt)} )
 
 

A= (int.(e^ln(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