Let y:R→Rsatisfy the initial value problem
y′(t)=1−y2(t),t∈R,y(0)=0
Then
(a) y(t1)=1 for some t1∈R.
(b) y(t) > −1 for all t ∈ R.
(c) y is strictly increasing in R.
(d) y is increasing in (0, 1) and decreasing in (1,∞).
Adv.
Consider the initial value problem (IVP) dydx=y2,y(0)=1,(x,y)∈R×R .
Then there exists a unique solution of the IVP on
(a) (−∞, ∞)
(b) (−∞, 1)
(c) (−2, 2)
(d) (−1,∞)
Consider the initial value problem (IVP) dydx=xy12,y(0)=0,(x,y)∈R×R . Then, which of the following are correct?
(a) The function f(x, y) = xy13 does not satisfy a Lipschitz condition with respect to y in any neighborhood of y = 0.
(b) There exists a unique solution for the IVP.
(c) There exists no solution for the IVP.
(d) There exists more than one solution of the IVP.
Letf:R→R be a continuous function with period p > 0. Then g(x)=∫x+pxf(t)dt is a
(a) constant function
(b) continuous function
(c) continuous function but not differentiable
(d) neither continuous nor differentiable
The solution of the differential equationdxdt=x2 with the initial condition x(0) = 1 will blow up as t tends to
(a) 1 (b) 2 (c) 1/2 (d) ∞
Let y1(x)andy2(x) be the solutions of the differential equation dydx=y+17 with initial conditions y1(0)=0,y2(0)=1. Then
(a) y1 and y2 will never intersect.
(b) y1 and y2 will intersect at x = 17.
(c) y1 and y2 will intersect at x = e.
(d) y1 and y2will intersect at x = 1.
The differential equation
dydx=60(y2)15;x>0,y(0)=0
has
(a) a unique solution.
(b) two solutions.
(c) no solution.
(d) infinite number of solutions.
Consider the initial value problem y′(t)=f(t)y(t),y(0)=1, where f:R→R is continuous. Then this initial value problem has
(a) indefinitely many solutions for some f.
(b) a unique solution in R.
(c) no solution in R for some f.
(d) a solution in the interval containing 0, but not on R for some f.
Consider the equation
dydx=(1+f2(t))y(t),y(0)=1:t≥0,
where f is a bounded continuous function on [0, ∞). Then
(a) The equation admits a unique solution y(t) and further Limt→∞y(t) exists and is finite
(b) The equations admits two linearly independent solutions
(c) This equation admits a bounded solution for which Limt→∞y(t) does not exist
(d) The equation admits a unique solution y(t) and further, Limt→∞y(t)=∞
Consider the differential equation
dydx=y2,(x,y)∈R×R
(a) all solutions of the differential equation are defined on (−∞,∞).
(b) no solution of the differential equation are defined on (−∞,∞).
(c) the solution of the differential equation satisfying the initial condition y(x0)=y0,y0>0,is defined on (−∞,x0+1y0)
(d) the solution of the differential equation satisfying the initial condition y(x0)=y0,y0>0, is defined on (x0−1y0,∞).
The initial value problem
˙x(t)=3x23,x(0)=0
in an interval around t = 0, has
(a) no solution
(b) a unique solution
(c) finitely many linearly independent solutions
(d) infinitely many linearly independent solutions
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