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

The first order system of equations equivalent to

\(x''+p(t)x'+q(t)x=0\)

is

\(\frac{dX}{dt} =A(t)X with \; X= {\begin{pmatrix} x_1\\x_2\end{pmatrix} },\)and \(\frac{dX}{dt}=\begin{pmatrix} x'_1\\x'_2 \end{pmatrix},\)

where A(t) is the \(2\times2 matrix\)

(a)  \(\begin{pmatrix} 1&0\\-p&-q \end{pmatrix}\)   (b)\(\begin{pmatrix}0&1\\-q&-p \end{pmatrix}\)(c)\(\begin{pmatrix} 0&1\\q&p \end{pmatrix}\)   (d) \(\begin{pmatrix}-p&-q \\1&0\end{pmatrix}\)

The function f(x, y) = x|y|,(x, y) ∈ R × R with respect to the y variable

(a) satisfies a Lipschitz condition on the strip −1 ≤ x ≤ 1, −∞ < y < ∞.

(b) satisfies a Lipschitz condition on the strip −∞ < x < ∞, −1 ≤ y ≤ 1.

(c) does not satisfy a Lipschitz condition on −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.

(d) does not satisfy a Lipschitz condition at any point on R × R.