The singular integral of the ODE 

$(xy'-y)^2 =x^2(x^2-y^2)$

is

(a) $y=xsinx$       (b)  $y=xsin(x+\frac{\pi}{4})$   (c)  $y=x$     (d)  $y=x+ \frac{\pi}{4}$

The initial value problem

$y'=2\sqrt{y}, y(0)=a,$

has

(a) a unique solution if a < 0,

(b) no solution if a > 0,

(c) infinitely many solutions if a = 0

(d) a unique solution if a ≥ 0

For the initial value problem

$\begin {cases} \frac{dy}{dx}=y^2+cos^2x, x>0,\\ y(0)=0 \end{cases}$

The largest interval of existence of the solution predicted by Picard’s theorem is:

(a) [0, 1] (b) [0, $\frac{1}{2}$ ] (c) [0, $\frac{1}{3}$ ] (d) [0, $\frac{1}{4}$ ]

Consider the ODE $y'=f(y(x))$. If f is an even function and y is an odd function, then

(a) $-y(-x)$is also a solution.

(b)$y(-x)$is also a solution.

(c) $-y(x)$ is also a solution.

(d) $y(x)y(-x)$ is also a solution.

Let $y:[o,\infty) \rightarrow [0,\infty)$be a continuously differentiable function satisfying 

$y(t)=y(0) +\int_0^t y(s)ds,$ for $t\ge0.$

Then

(a)   $y^2(t)=y^2(0)=\int_0^t y^2(s)ds.$

(b)   $y^2(t)=y^2(0)+2\int_0^t y^2(s)ds.$

(c)   $y^2(t)=y^2(0)+\int_0^t y(s)ds.$

(d)   $y^2(t)=y^2(0)+\left(\int_0^ty(s)ds\right)^2 +2y(0)\int_0^ty(s)ds.$

Let u(t) be a continuously differentiable function taking non-negative values for t > 0 and satisfying $u'(t)=4u^\frac{3}{4}(t);u(0)=0.$Then

(a)  u(t)=0.

(b)  $u(t)=t^4$

(c)   $u(t)=\begin{cases} 0\;\;\; for\, 0<t<1,\\ (t-1)^4 \;\;for \;t\ge1 \end{cases}$

(d)   $u(t)=\begin{cases} 0\;\;\; for\, 0<t<10,\\ (t-10)^4 \;\;for \;t\ge10 \end{cases}$

Consider the integral equation

$y(x)=x^3=\int_0^xsin(x-t)y(t)dt,x\in[0,\pi].$

Then the value of y(1) is

(a) $\frac{19}{20}$       (b) 1       (c) $\frac{17}{20}$         (d) $\frac{21}{20}$

Let (x(t), y(t)) satisfy the system of ODEs

$\frac{dx}{dt}=-x+ty,\\ \frac{dy}{dt}=tx-y.$

If $(x_1(t),y_1(t)) \;and\; (x_2(t),y_2(t))$ are two solutions and$\Phi(t)=x_1(t)y_2(t)-x_2(t)y_1(t),$, then $\frac{d\Phi}{dt}$is equal to

a) $-2\Phi$      (b) $2\Phi$       (c) $-\Phi$         (d) $\Phi$

Consider the initial value problem

$y'(t)=f(y(t)),y(0)=a\in \mathbb{R}$

 Which of the following statements are necessarily true?

(a) There exists a continuous function$f:\mathbb{R}\rightarrow \mathbb{R}$  and $a\in\mathbb{R}$ such that the above problem does not have a solution in any neighbourhood of 0.

(b) The problem has a unique solution for every $a\in\mathbb{R}$, when f is Lipschitz continuous.

(c) When is twice continuously differentiable, the maximal interval of existence for the above initial value problem is $\mathbb{R}$.

(d) The maximal interval of existence for the above problem is$\mathbb{R}$, when is bounded and continuously differentiable.

Suppose x : [0,∞) → [0,∞) is continuous and x(0) = 0. If 

$x(t)\le2+\int_0^tx(s)ds,for \;all\:t\ge 0$, for all t ≥ 0,

then which of the following is true?

(a) x( √ 2) ∈ [0, 2]

(b) x( √ 2) ∈ [0, $2e^\sqrt2$]

(c) x( √ 2) ∈ [$\frac{5}{\sqrt2},\frac{7}{\sqrt2}$]

(d) x( √ 2) ∈ [10,∞)

Consider the ODE

$y'(t)=-y^3+y^2+2y,subject\:to\:y(0)=y_0\in(0,2).$

Then $Lim_{t\rightarrow\infty}y(t)$ belongs to

(a) {−1, 0}    (b) {−1, 2}       (c) {0, 2}        (d) {0, +∞}