Let u(t) be a continuously differentiable function taking non-negative values for t > 0 satisfying $u'(t)=3u(t)^{\frac{2}{3}}$ and u(0) = 0. Which of the following are possible solutions of the above equation?

(a) u(t) = 0

(b) u(t) = $t^3$

(c) $u(t)= \begin{cases} 0 \; for \; 0<t<1 \\ (t-1)^3 for \;t \ge 3 \end{cases}$

(d) $u(t)=\begin{cases} 0 \; for \; 0<t<1\\ (t-3)^3\; for \;t\ge3 \end{cases}$

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  $(x_0-\frac{1}{y_0},\infty)$.