1
GATE ME 2014 Set 3
MCQ (Single Correct Answer)
+2
-0.6
Consider two solutions $$\,x\left( t \right)\,\,\,\, = \,\,\,{x_1}\left( t \right)\,\,$$ and $$x\left( t \right)\,\,\,\, = \,\,\,{x_2}\left( t \right)\,\,$$ of the differential equation
$$\,\,{{{d^2}x\left( t \right)} \over {d{t^2}}} + x\left( t \right) = 0,t > 0,\,\,$$ such that
$$\,{x_1}\left( 0 \right) = 1,{\left. {{{d{x_1}\left( t \right)} \over {dt}}} \right|_{t = 0}} = 0,$$ $$\,\,\,\,{x_2}\left( 0 \right) = 0,{\left. {{{d{x_2}\left( t \right)} \over {dt}}} \right|_{t = 0}} = 1$$

The wronskian $$\,w\left( t \right) = \left| {{{\matrix{ {{x_1}\left( t \right)} \cr {d{x_1}\left( t \right)} \cr } } \over {dt}}} \right.\left. {{{\matrix{ {{x_2}\left( t \right)} \cr {d{x_2}\left( t \right)} \cr } } \over {dt}}} \right|$$ at $$\,\,t = \pi /2$$

A
$$1$$
B
$$-1$$
C
$$0$$
D
$$\pi /2$$
2
GATE ME 2014 Set 1
Numerical
+2
-0
If $$\,y = f\left( x \right)\,\,$$ is the solution of $${{{d^2}y} \over {d{x^2}}} = 0$$ with the boundary conditions $$y=5$$ at $$x=0,$$ and $$\,{{dy} \over {dx}} = 2$$ at $$x=10,$$ $$f(15)=$$_______.
Your input ____
3
GATE ME 2014 Set 1
MCQ (Single Correct Answer)
+2
-0.6
The matrix form of the linear system $${{dx} \over {dt}} = 3x - 5y$$ and $$\,{{dy} \over {dt}} = 4x + 8y\,\,$$ is
A
$${d \over {dt}}\left\{ {\matrix{ x \cr y \cr } } \right\} = \left[ {\matrix{ 3 & { - 5} \cr 4 & 8 \cr } } \right]\left\{ {\matrix{ x \cr y \cr } } \right\}$$
B
$${d \over {dt}}\left\{ {\matrix{ x \cr y \cr } } \right\} = \left[ {\matrix{ 3 & 8 \cr 4 & { - 5} \cr } } \right]\left\{ {\matrix{ x \cr y \cr } } \right\}$$
C
$${d \over {dt}}\left\{ {\matrix{ x \cr y \cr } } \right\} = \left[ {\matrix{ 4 & { - 5} \cr 3 & 8 \cr } } \right]\left\{ {\matrix{ x \cr y \cr } } \right\}$$
D
$${d \over {dt}}\left\{ {\matrix{ x \cr y \cr } } \right\} = \left[ {\matrix{ 4 & 8 \cr 3 & { - 5} \cr } } \right]\left\{ {\matrix{ x \cr y \cr } } \right\}$$
4
GATE ME 2013
MCQ (Single Correct Answer)
+2
-0.6
The solution to the differential equation $$\,{{{d^2}u} \over {d{x^2}}} - k{{du} \over {dx}} = 0\,\,\,$$ where $$'k'$$ is a constant, subjected to the boundary conditions $$\,\,u\left( 0 \right) = 0\,\,$$ and $$\,\,\,u\left( L \right) = U,\,\,$$ is
A
$$u = U{x \over L}$$
B
$$u = U\left( {{{1 - {e^{kx}}} \over {1 - {e^{kL}}}}} \right)$$
C
$$u = U\left( {{{1 - {e^{ - kx}}} \over {1 - {e^{ - kL}}}}} \right)$$
D
$$u = U\left( {{{1 + {e^{kx}}} \over {1 + {e^{kL}}}}} \right)$$
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