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Math  /  Calculus

Question(12) Let y1y_{1} and y2y_{2} be two solutions of the DE t2y′′−t(t+1)y′+y=0,t>0.t^{2} y^{\prime \prime}-t(t+1) y^{\prime}+y=0, \quad t>0 .
If W(y1,y2)(2)=2e2W\left(y_{1}, y_{2}\right)(2)=2 \mathrm{e}^{2}, then W(y1,y2)(−1)=W\left(y_{1}, y_{2}\right)(-1)= (a) -e (b) e−1\mathrm{e}^{-1} (c) e2e^{2} (d) −e−1-\mathrm{e}^{-1} (e) ee

Studdy Solution
Evaluate the Wronskian at t=−1 t = -1 :
W(−1)=e−1⋅∣−1∣=e−1 W(-1) = e^{-1} \cdot |-1| = e^{-1}
The value of the Wronskian at t=−1 t = -1 is e−1 \boxed{e^{-1}} .

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