Consider two solutions x(t) = x 1 (t) and x(t) = x 2 (t) of the d
Consider two solutions x(t) = x1(t) and x(t) = x2(t) of the differential equation \(\frac{{{d^2}x\left( t \right)}}{{d{t^2}}} + x\left( t \right) = 0,\;t > 0\), such that x1(0) = 1, \({\left. {\frac{{d{x_1}\left( t \right)}}{{dt}}} \right|_{t = 0}} = 0,\;{x_2}\left( 0 \right) = 0,{\left. {\frac{{d{x_2}\left( t \right)}}{{dt}}} \right|_{t = 0}} = 1\)
The Wronskian \(W\left( t \right) = \left| {\begin{array}{*{20}{c}} {{x_1}\left( t \right)}&{{x_2}\left( t \right)}\\ {\frac{{d{x_1}\left( t \right)}}{{dt}}}&{\frac{{d{x_2}\left( t \right)}}{{dt}}} \end{array}} \right|\) at \(t = \frac{\pi}{2}\) is
A. 1
B. -1
C. 0
D. π/2
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Right Answer is: A
SOLUTION
Concept:
Given differential equation is
\(\frac{{{d^2}x}}{{d{t^2}}} + x = 0\)
∴ Auxillary equation is
m2 + 1 = 1
⇒ m = ± i
∴ General solution is
x = C1 cost + C2 sint
\(\Rightarrow \dot x = - {C_1}\sin t + {C_2}\cos t\)
For \({x_1}\left( 0 \right) = 1\;\& \frac{{d{x_1}\left( 0 \right)}}{{dt}} = 0\)
⇒ x1 = cos t
For \({x_2}\left( 0 \right) = 0\;\& \frac{{d{x_2}\left( 0 \right)}}{{dt}} = 1\)
⇒ x2 = sin t
\(W = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{x_2}}\\ {\frac{{d{x_1}}}{{dt}}}&{\frac{{d{x_2}}}{{dt}}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\cos t}&{\sin t}\\ { - \sin t}&{\cos t} \end{array}} \right|\)
= cos2t + sin2t = 1