Exercise · ZW-WRONSKIAN-004 · ★☆☆ Basic
Compute \(W(e^{t},2e^{t})\) and interpret the result.
Choose one option.
Show answer
- D. \(W=0\) for every \(t\): they are linearly dependent
Explanation
\(W=e^t\cdot2e^t-e^t\cdot2e^t=0\). The function \(2e^t\) is twice \(e^t\), so they are dependent.
Review the concept: Wronskian
The Wronskian \(W\) of two differentiable functions \(y_1\) and \(y_2\) is the determinant of the matrix formed by them and their first derivatives.
\[W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\y_1'&y_2'\end{vmatrix}=y_1y_2'-y_1'y_2\]
More practice problems: wronskian
- ZW-WRONSKIAN-001 Compute the Wronskian W(e^2t,e^3t).
- ZW-WRONSKIAN-002 Compute W(t,t^2).
- ZW-WRONSKIAN-003 Compute W(cos t,sin t).
- ZW-WRONSKIAN-005 Compute W(e^-t,te^-t).
- ZW-WRONSKIAN-006 Compute W(sin 2t,cos 2t).
- ZW-WRONSKIAN-007 For t>0, compute W(t, tln t).
- ZW-WRONSKIAN-008 Let y_1,y_2 be solutions of y''+py'+qy=0, with p,q continuous, and suppose W(y_1,y_2)(t_0)≠0 at a p…
- ZW-WRONSKIAN-009 Let y_1,y_2 be solutions of y''+py'+qy=0 with p,q continuous, and W(y_1,y_2)(t_0)=0 at some t_0. Wh…