MATHEMATICAL PROBLEM PAGE Directed
by EDWARD L. HAENISCH
Montana State College, Bozeman, Montana
A
N ORDINARY differential equation is a relation between the variables, x and y, and one or more of the derivatives, d-,y d2y etc. Theorder dx dx2 of an equation is the order of the highest derivative which appears in it. The degree is the power to which this highest derivative is raised after the equation has been cleared of fractions and radicals. The general solution of a differential equation contains a number of arbitrary independent constants equal to the order of the equation. In most applications of differential equations to physical phenomena these constants can be evaluated for the conditions of the problem a t hand. First-order, first-degree equations have the form:
-.
+
Fdx,~f ) F & , Y ) ( ~ Y / ~ x= ) 0
Ndy = 0, where M = Fl(x,y) and N = or Mdx There are several methods for the solution F2(x,y). of such equations. This month we only two types: (1) The variables are separable. The equation is either in the form that M is a function o f x alone and N a function of y alone or i t can be put in this form by slight manipulation. The solution is obtained by simple integration:
"Higher mathematics for chemical students," 3rd ed., D. Van Nostrand Co., Inc., New York City, p. 214.
PARTINGTON,
PROBLEMS
1. Obtain a differential equation free from arbitrary 6% is a general constants for which y = ax8 solution. 2. Solve: yxdx xyzdy = 0 (x2 - yxZ) (dy/dx) y2 xy2 = 0 3. Solve: xaydx - (x3 ya)dy = 0 4. Solve: ( 3 y - 7 x 7)dx (7y 3x 3)dy = 0 5. For the adiabatic expansion of a perfect gas (OV = RT) the differential equation is fidV = :&T wheie c, is the heat capacity a t constant volume and may be considered constant. Solve the equation. Show also that pVr = const. R. where 7 = q/c, and c*=c, 6. Solve the differential equation for the velocity of a bimolecular reaction:
+
+
+ + + + + + + +
dx/dt = k(a - x) (b - x) Evaluate the arbitrary constant from the fact that a t t = 0, x = 0. SOLUTIONS TO JANUARY PROBLEMS
(2) The equation is homogeneous. An equation is homogeneous with respect to x and y if the sum of the powers of x and y in each term is a constant. Thus, a homogeneous equation can contain no constant term. Such equations may be transformed into ones which are solvable by separation of the variables by the substitutions of y = ux and dy = vdx f xdxC
a2u . a 4 asu axu - is negative and - - - b ra axP ay* axay and . y = llsisamaxiis -ereater than zero. Therefore z = . mum paint in the given function.
REFERENCES DANIELS:
MELLOR:
Under these conditions
187 370
-
* An equation of the first degree with respect to y and r of the type: (a'%+ b'y +c')dy- (ax + b y +c)dl: = 0 can be transformed into the homogeneous form by the substitution of x - u h, y = v k , d r = du and d y = du; h and k are chosen so that ah bk c = 0 and a'h b'k c' = 0.
+ + ++
1 . a u / a x = 3x9' - 4s8y' - 3x2ya = x'yP(3 - 4% - 3 y ) 64/bx" 66y2 - 12zqya - 6xya = z y a ( 6 - 12x - 6 y ) a~ / b y = 2 x 9 - 2 x 4 - 3 ~ =9~ "~( 2- 2~ - 3 y ) b 2 u / b y 2 = 2 r S - 2%' - 6xSy = x3(2 - 22 - 6 y ) b % / ( b x b y ) = 6 2 4 - 8x8y - 9x2ys = xSy(6 - 82 - Q y ) d u / b x = a u / b y = 0 for the solution of the twosimultaneous equations: 3 - 4x - 3 y = 0 and 2 - 2% - 3 y = 0.
+ +
.
2 . Let the number C be divided into three parts, x, y, and ( C x - y ) . The problem requires the relationship between xand y when the function u = xy(C - x - y) is at amaximum.
have been inexact. I n this case if-y = .xl is adopted as the equation of a path passing through the points 0,O and 1.1, Function may be a maximum or minimum for the solution of the simultaneous equations:
the integral reduces to
J: - x2dx = ".
IF y = xa is
J
adopted a s thepath, the integral is - ~ X U X = -L/~ C-2%-y=o C-x-2y=o 6. U is a function of three variables, 9 , V , T , any two of which The solution is x = C/3 and Y = C/3. For these values of x may be taken as the independent variables. aPu $2; a2u ' a2* . and y, -1s negative and ~spreaterthanzero. dU = ( b U / h T ) v d T (bU/bV)TdV axa axs byZ a ~ a y Therefore the number C should be divided into three eoual This is substituted into the expression for dQ and we have: parts if the product of these parts is to be a maximum dQ = ( d U l b T ) ~ d T K b U / a V h + P I dV
.
-.
+
+
a2u ap b2u Testing this for exactness we obtain: -+ -- - -. bVbT bTbV bT The given function has no maxima or minima.
dQ is an inexact differential
4. a M / a y = 8 eicos y; bN/bx = e" cos y . The differential is exact and u = e' sin y C. b M / & = nV"-1; b N / b p = nVn-1. The differential is exact and u = p P C aM/ay = ---ax f Zy; b N / & = y - 2 m . The differential is inexact.
+
+
aM aN 5. - = 2%; - = 2%. The differential is exact and u
a~
a%
=
xay.
+ ( I / T ) [ ( a U / a V )+ PI
dQ/T = ( l / T ) ( b U / b T )
dV
Testing this for exactness we obtain:
Since order of differentiation is immaterial and since under the conditions of the problem b U / b V = Q, the above becomes:
( l / T ) ( b P / a T )= P I P ap/aT = R / V As the next part of the problem appears in THE JOURNAL it is also exact and the value is independent of the path. If the problem had read
(ydz
- xdy) the differential would
R / ( V T ) = p/T2, or pV = RT. Therefore we have proved that for a perfect gas dQ is an ineaact differential and dQ/T is an exact differential.
