Answers to Selected Exercises 545
- The differential equation is nonlinear and f(x, y) = 3xy^113 is continu-
ous on the entire plane, so a solution exists until x ---+ -oo, x ---+ +oo,
y---+ -oo, or y ---+ +oo. The function fy(x, y) = xy-^2!^3 is continuous
for y-/= 0.
a ., b., c., and e. The solution exists and is unique at least until y = 0
where multiple solutions may exist.
d. The solution exists, but it may not be unique.
a. The unique, explicit solution on ( -oo, oo) is
y = (x2 + ( 9 / 4 )1/3 _ l)3/2.
b. The unique, explicit solution on (-oo, 0) is y = - x^3.
c. The unique, explicit solut ion on (-oo, -Jl - (1/4)^113 ) is
y = (x2 + (l/ 4 )1/3 _ l)3/2_
d. The solution is not unique. The function y = 0 is a solution on
(-00,00) and y = ±(x^2 -1)^312 a re solut ions on (-oo,-1).
e. The unique, explicit solution on (-oo, O) is y = x^3.
- The differential equation is nonlinear and f(x, y) = y/(y - x) and
fy(x, y) = -x/(y - x)^2 are both continuous for x-/= y.
a ., c., and d. Thus, the solution exists and is unique until the solution
reaches the line y = x.
b. There is no solution, because the differential equation is undefined
at (1, 1).
a. The unique solution on (0, oo) is y = 2x.
c. The unique solution on (0, oo) is y = 0.
d. The unique solution on ( -oo, oo) is y = x - -J x^2 + 3.
- The differential equation is nonlinear. The function f(x, y) = xJl - y^2
is real and continuous for -1 :::; y :::; 1, so a solution exists so long as it
remains in {(x,y)l -1:::; y:::; l}. Since fy(x,y) = -xy/~ is real
and continuous for -1 < y < 1, the solution exists and is unique so long
as it remains in {(x,y)l-1:::; y:::; l}. If the solution reaches y = 1 or
y = -1, it may no longer be unique.
a. The solution is not unique. Both y = 1 and y = sin( x
2
;t7r) are
solutions on (-00,00).
2
b. The unique solution on ( -oo, oo) is y = sin( x 2 + arcsin .9).
2
c. The unique solution on ( -oo, oo) is y = sin( x 2 + i).
2
d. The unique solution on (-oo, oo) is y = sin("; ).
