1
GATE CE 2005
MCQ (Single Correct Answer)
+2
-0.6
The solution $${{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} + 17y = 0;$$ $$y\left( 0 \right) = 1,{\left( {{{d\,y} \over {d\,x}}} \right)_{x = {\raise0.5ex\hbox{$\scriptstyle \pi $} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}}} = 0\,\,$$ in the range $$0 < x < {\pi \over 4}$$ is given by
A
$${e^{ - x}}\left[ {\cos \,4x + {1 \over 4}\sin \,4x} \right]$$
B
$${e^x}\left[ {\cos \,4x - {1 \over 4}\sin \,4x} \right]$$
C
$${e^{ - 4x}}\left[ {\cos \,4x - {1 \over 4}\sin \,x} \right]$$
D
$${e^{ - 4x}}\left[ {\cos \,4x - {1 \over 4}\sin \,4x} \right]$$
2
GATE CE 2004
MCQ (Single Correct Answer)
+2
-0.6
Biotransformation of an organic compound having concentration $$(x)$$ can be modeled using an ordinary differential equation $$\,{{d\,x} \over {dt}} + k\,{x^2} = 0,$$ where $$k$$ is the reaction rate constant. If $$x=a$$ at $$t=0$$ then solution of the equation is
A
$$x = a\,{e^{ - kt}}$$
B
$$\,{1 \over x} = {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle a$}} + k\,t$$
C
$$x = a\left( {1 - {e^{ - kt}}} \right)$$
D
$$x = a\, + k\,t$$
3
GATE CE 2001
MCQ (Single Correct Answer)
+2
-0.6
The solution for the following differential equation with boundary conditions $$y(0)=2$$ and $$\,\,{y^1}\left( 1 \right) = - 3$$ is where $${{{d^2}y} \over {d{x^2}}} = 3x - 2$$
A
$$y = {{{x^3}} \over 3} - {{{x^2}} \over 2} = 3x - 2$$
B
$$y = 3{x^3} - {{{x^2}} \over 2} - 5x + 2$$
C
$$y = {{{x^3}} \over 2} - {x^2} - 5{x \over 2} + 2$$
D
$$y = {x^3} - {{{x^2}} \over 2} + 5x + {3 \over 2}$$
4
GATE CE 1998
Subjective
+2
-0
Solve $${{{d^4}y} \over {d{x^4}}} - y = 15\,\cos \,\,2x$$
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