The solution of the differential equation $\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} + 2\frac{{{\rm{dy}}}}{{{\rm{dt}}}} + {\rm{y}} = 0$ with ${\rm{y}}\left( 0 \right) = {\rm{\;y'}}\left( 0 \right){\rm{\;}} = {\rm{\;}}1{\rm{\;}}$is

Question

The solution of the differential equation $\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} + 2\frac{{{\rm{dy}}}}{{{\rm{dt}}}} + {\rm{y}} = 0$ with ${\rm{y}}\left( 0 \right) = {\rm{\;y'}}\left( 0 \right){\rm{\;}} = {\rm{\;}}1{\rm{\;}}$is

TuteeHUB · Accepted Answer

$\left( {2{\rm{\;}} + {\rm{\;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}$

TuteeHUB · Answer

$\left( {2{\rm{\;}} - {\rm{\;t}}} \right){{\rm{e}}^{\rm{t}}}$

TuteeHUB · Answer

$\left( {1{\rm{\;}} + {\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{ - {\rm{t}}}}$

TuteeHUB · Answer

$\left( {1{\rm{\;}}-{\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{\rm{t}}}$

1.	The solution of the differential equation \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} + 2\frac{{{\rm{dy}}}}{{{\rm{dt}}}} + {\rm{y}} = 0\) with \({\rm{y}}\left( 0 \right) = {\rm{\;y'}}\left( 0 \right){\rm{\;}} = {\rm{\;}}1{\rm{\;}}\)is
A.	\(\left( {2{\rm{\;}} - {\rm{\;t}}} \right){{\rm{e}}^{\rm{t}}}\)
B.	\(\left( {1{\rm{\;}} + {\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{ - {\rm{t}}}}\)
C.	\(\left( {2{\rm{\;}} + {\rm{\;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}\)
D.	\(\left( {1{\rm{\;}}-{\rm{\;}}2{\rm{t}}} \right){{\rm{e}}^{\rm{t}}}\)
Answer» C. \(\left( {2{\rm{\;}} + {\rm{\;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}\)

The solution of the differential equation \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} + 2\frac{{{\rm{dy}}}}{{{\rm{dt}}}} + {\rm{y}} = 0\) with \({\rm{y}}\left( 0 \right) = {\rm{\;y'}}\left( 0 \right){\rm{\;}} = {\rm{\;}}1{\rm{\;}}\)is

Discussion

No Comment Found

Related MCQs

Reply to Comment