Let \(y:\mathbb{R}\rightarrow \mathbb{R}\)satisfy the initial value problem

\(y'(t)=1-y^2(t),t\in\mathbb{R},y(0)=0\)

Then

(a) \(y(t_1)=1\) for some \(t_1\in\mathbb{R}\).

(b) y(t) > −1 for all t ∈ \(\mathbb{R}\).

(c) y is strictly increasing in \(\mathbb{R}\).

(d) y is increasing in (0, 1) and decreasing in (1,∞).

Consider the initial value problem (IVP) \(\frac{dy}{dx}=y^2,y(0)=1,(x,y)\in \mathbb{R} \times \mathbb{R}\) .

Then there exists a unique solution of the IVP on 

(a) (−∞, ∞)

(b) (−∞, 1)

(c) (−2, 2)

(d) (−1,∞)

Consider the initial value problem (IVP) \(\frac{dy}{dx}=xy^{\frac{1}{2}},y(0)=0,(x,y)\in \mathbb{R} \times \mathbb{R}\) . Then, which of the following are correct?

(a) The function f(x, y) = \(xy^{\frac{1}{3}}\) 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.

Let\(f:\mathbb{R} \rightarrow \mathbb{R}\) be a continuous function with period p > 0. Then \(g(x)=\int_x^{x+p}f(t)dt\) is a

(a) constant function

(b) continuous function

(c) continuous function but not differentiable

(d) neither continuous nor differentiable

 The solution of the differential equation\(\frac{dx}{dt}=x^2\) with the initial condition x(0) = 1 will blow up as t tends to

(a) 1       (b) 2        (c) 1/2        (d) ∞

Let \(y_1(x)and \;y_2(x)\) be the solutions of the differential equation \(\frac{dy}{dx}=y+17\) with initial conditions \(y_1(0)=0,y_2(0)=1\). Then

(a) \(y_1\) and \(y_2\) will never intersect.

(b) \(y_1\) and \(y_2\) will intersect at x = 17.

(c) \(y_1\) and \(y_2\) will intersect at x = e.

(d) \(y_1\) and \(y_2\)will intersect at x = 1.

The differential equation

\( \frac{dy}{dx}=60(y^2)^{\frac{1}{5}};x>0,y(0)=0\)

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 \(f:\mathbb{R}\rightarrow\mathbb{R}\) is continuous. Then this initial value problem has

(a) indefinitely many solutions for some \(f\).

(b) a unique solution in \(\mathbb{R}\).

(c) no solution in \(\mathbb{R}\) for some \(f\).

(d) a solution in the interval containing 0, but not on \(\mathbb{R}\) for some \(f\).

Consider the equation

\(\frac{dy}{dx}=(1+f^2(t))y(t),y(0)=1:t\ge 0,\)

where f is a bounded continuous function on [0, ∞). Then

(a) The equation admits a unique solution y(t) and further \(Lim_{t\rightarrow \infty}y(t)\) exists and is finite

(b) The equations admits two linearly independent solutions

(c) This equation admits a bounded solution for which \(Lim_{t\rightarrow \infty}y(t)\) does not exist

(d) The equation admits a unique solution y(t) and further, \(Lim_{t\rightarrow \infty}y(t)=\infty\)

Consider the differential equation

\(\frac{dy}{dx}=y^2,(x,y)\in\mathbb{R}\times\mathbb{R}\)

Then

(a) all solutions of the differential equation are defined on \((-\infty,\infty)\).

(b) no solution of the differential equation are defined on \((-\infty,\infty)\).

(c) the solution of the differential equation satisfying the initial condition \(y(x_0)=y_0,y_0>0,\)is defined on \((-\infty,x_0+\frac{1}{y_0})\)

(d) the solution of the differential equation satisfying the initial condition \(y(x_0)=y_0,y_0>0\), is defined on  \((x_0-\frac{1}{y_0},\infty)\).

The initial value problem

\(\dot{\mathbf{x}}(t)=3x^{\frac{2}{3}},x(0)=0\)

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