1
GATE CE 2010
MCQ (Single Correct Answer)
+2
-0.6
The solution to the ordinary differential equation $${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} - 6y = 0\,\,\,$$ is
A
$$y = {C_1}\,{e^{3x}} + {C_2}\,{e^{ - 2x}}$$
B
$$y = {C_1}\,{e^{3x}} + {C_2}\,{e^{2x}}$$
C
$$y = {C_1}\,{e^{ - 3x}} + {C_2}\,{e^{2x}}$$
D
$$y = {C_1}\,{e^{ - 3x}} + {C_2}\,{e^{ - 2x}}$$
2
GATE CE 2007
MCQ (Single Correct Answer)
+2
-0.6
The solution for the differential equation $$\,{{d\,y} \over {d\,x}} = {x^2}\,y$$ with the condition that $$y=1$$ at $$x=0$$ is
A
$$y = {e^{{1 \over {2x}}}}$$
B
$$\ln \left( y \right) = {{{x^3}} \over 3} + 4$$
C
$$\ln \left( y \right) = {{{x^2}} \over 2}$$
D
$$y = {e^{{{{x^3}} \over 3}}}$$
3
GATE CE 2005
MCQ (Single Correct Answer)
+2
-0.6
Transformation to linear form by substituting $$v = {y^{1 - n}}$$ of the equation $${{dy} \over {dt}} + p\left( t \right)y = q\left( t \right){y^n},\,\,n > 0$$ will be
A
$${{dv} \over {dt}} + \left( {1 - n} \right)pv = \left( {1 - n} \right)q$$
B
$${{dv} \over {dt}} + \left( {1 + n} \right)pv = \left( {1 + n} \right)q$$
C
$${{dv} \over {dt}} + \left( {1 + n} \right)pv = \left( {1 - n} \right)q$$
D
$${{dv} \over {dt}} + \left( {1 - n} \right)pv = \left( {1 + n} \right)q$$
4
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
$${e^{ - 4x}}\left[ {\cos \,4x - {1 \over 4}\sin \,4x} \right]$$
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