Exercise · ZW-WRONSKIAN-010 · ★★★ Advanced
Two solutions of \(y''+3y'+2y=0\) have \(W(0)=2\). What is \(W(t)\)?
Choose one option.
Show answer
- B. \(2e^{-3t}\)
Explanation
Abel's formula gives \(W'=-pW=-3W\), with \(p=3\), so \(W(t)=2e^{-3t}\). Check with \(e^{-t},e^{-2t}\): \(W=-e^{-3t}\), proportional.
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-004 Compute W(e^t,2e^t) and interpret the result.
- 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…