If the largest eigen value of the matrix\(\left[\begin{array}{ccc} 5&-2&0 \\ 1&2&-3 \\ 1&-2&4\end{array}\right]\)is 6 and corresponding eigen vector is\(\left[\begin{array}{ccc} 1 \\ -1/2 \\ 1\end{array}\right]\),then find the subdominant eigen value.

Find the eigen values of following matrix:

\(\left[\begin{array}{ccc} 1&-1&0&0&0&0 \\ 1&1&-1&0&0&0 \\ 0&1&1&-1&0&0\\0&0&1&1&-1&0\\0&0&0&1&1&-1\\0&0&0&0&1&1\end{array}\right]\)

Let y(x) be a continuous solution of the initial value problem

y' + 2y = f(x), y(0) = 0, where \( f(x) = \begin{cases} 1, & \quad \text{if } 0 \le x\le1,\\ 0, & \quad \text{if } x>1 \end{cases} \)

Then y\(( \frac{3}{2})\) is equal to

(a)  \(( \frac{sinh(1)}{e^3})\)  (b)   \(( \frac{cosh(1)}{e^3})\)   (c)  \(( \frac{sinh(1)}{e^2})\)  (d)   \(( \frac{cosh(1)}{e^2})\)

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) \(\Phi\)