Karl's Calculus Forum: November/December 2004

Reply to Gordon: You didn't give any initial condition, so I'll state one here: y(0) = A. This one is easy enough to find a closed solution for. The function multiplying y is x². You integrate that and take the exponential of the negative of the result. When you apply the initial condition you find

   y(x)  =  Ae-x3/3

You can also use the method of finding the powers one by one. Your equation is equivalent to y' = x²y. So you have

   y(0)  =  A     initial condition given
   y'(0) =  x2y  =  0  because x=0
   y"(0) =  2xy + x2y'  =  0  for the same reason
   y'"(0) = 2y + 4xy' + x2y"  =  2A

  dV dh    dV
        =    
  dh dt    dt

equation 1:     2x2y + x2 = c       (where c is a constant)
equation 2:     x + y = 1

hence y = 1 - x

  2x2(1 - x) + x2  =  c

  3x2 - 2x3  =  c

  3x2 - 2x3 - c  =  0

Problem 2:

equation 1:     6x3y2 + 3x3y + x3 = c       (where c is a constant)
equation 2:     x + y = 1

equation 1:     2x2y + x2 = c       (where c is a constant)
equation 2:     x + y = 1

equation 1:     6x3y2 + 3x3y + x3 = c       (where c is a constant)
equation 2:     x + y = 1

x 1

dt t

The log of 2x will be that same integral taken from 1 to 2x. Now break that integral into two pieces:

ln(2x) =

2 1

dt + t

2x 2

dt t

Now look at the 1/x curve from 2 to 2x. Isn't it just half of the same curve taken from 1 to x, except stretched out? Use that fact to evaluate the second integral in the sum in terms of ln(x). If you let u = 2t, and du = 2dt as a substitution in the second integral and substitute the limits as well, it all works out. Observe that the first integral in the sum is, by definition, ln(2). See if you can finish it now.
Karl <Click to Send Email to Karl>
USA - Tuesday, December 07, 2004 at 19:09:39 (EST)

From here you can proceed two ways. One is to recognize that the solution will be in the form of y = (ax² + bx + c)e^x + B, then plug this back into the original equation to solve for a, b, and c. Actually both c and B will be undetermined, and you determine them from the initial conditions given in the problem.

The other (more general) method is by variation of parameters. To do this you replace A and B in the homogeneous solution with two unknown functions, u(x) and v(x). So you have the solution, y = u(x)e^x + v(x). The trick now is to find u and v. To do this we solve the following two equations simultaneously:

   u'(x)ex + v'(x)  =  0

   u'(x)(d/dx)ex + v'(x)(d/dx)1  =  xex

   u'(x)ex  =  xex

   u'(x)  =  x

   u(x)  =  (1/2)x2 + C

I'm asked to solve y"-y'=xex, with initial conditions y(0)=2 and y'(0)=1. This is  a second order homogeneous differential equation, so I try to subsitute y=(Ax+B)*ex. After substituting the second derivative and first derivative of my y, I found that I'm left with Aex=xex. Can someone tell me what I did wrong?

   a2 + b2 = c2

  2a da/dt  +  2b db/dt  =  2c dc/dt

                     (ax1 + bx2)(x1 dc + x2 de)
            G  =     ---------------------
                     (a + c)x1 + (b + e)x2

where x1 + x2 = X.

   g(x)  =  x

   f(g(x))  =  3g(x) + 2  =  x

  3g + 2  =  x

   V  =  (1/3)πr2h

   A  =  πr2

   dA/dt  =  2πr dr/dt

For part b), you need to take the time-derivative of the volume equation using the product and chain rules:

   dV/dt  =  (1/3)π(2rh dr/dt  +  r2 dh/dt)

For part c), divide dA/dt by dh/dt
Karl <Click to Send Email to Karl>
USA - Monday, November 22, 2004 at 00:16:53 (EST)

   V  =  32000 cm  =  L2H

   A  =  4LH + L2

  H  =  V/L2

   A  =  4L(V/L2) + L2  =  4V/L + L2

  dA
      =  0  =  -4V/L2 + 2L
  dL

    I cant figure out this problem.
    a box with a square base and open top must have a volume of 32,000cm^2. find the demensions of the
box that minimize the amount of material used.

    is this a square box or what, i am having trouble with it.