Exercise · ZW-SECOND-ORDER-LINEAR-EQUATION-005 · ★☆☆ Basic
If \(y=e^{rt}\) solves \(y''-5y'+6y=0\), what values can \(r\) take?
Choose one option.
Show answer
- A. \(2\) or \(3\)
Explanation
Substituting gives \(r^2-5r+6=0\), that is \((r-2)(r-3)=0\).
Review the concept: Second-order linear equation
It is a second-order differential equation, linear in the unknown \(y\) and its derivatives, whose coefficients \(p\), \(q\) and term \(g\) depend only on \(t\).
\[y''+p(t)\,y'+q(t)\,y=g(t)\]
More practice problems: second-order linear equation
- ZW-SECOND-ORDER-LINEAR-EQUATION-001 Which of these equations is second-order linear?
- ZW-SECOND-ORDER-LINEAR-EQUATION-002 In y''+tfrac2t y'-3y=e^t, what is q(t)?
- ZW-SECOND-ORDER-LINEAR-EQUATION-003 Which of these equations is homogeneous (g=0)?
- ZW-SECOND-ORDER-LINEAR-EQUATION-004 Which function solves y''+3y'+2y=0?
- ZW-SECOND-ORDER-LINEAR-EQUATION-006 For k>0, which value makes y=sin (kt) solve y''+16y=0?
- ZW-SECOND-ORDER-LINEAR-EQUATION-007 The function y=t^2 solves y''+y=g(t). What is g(t)?
- ZW-SECOND-ORDER-LINEAR-EQUATION-008 What is the constant solution of y''+4y'+5y=10?
- ZW-SECOND-ORDER-LINEAR-EQUATION-009 Why is y''+y y'=0 not linear?
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