Exercise · ZW-CHARACTERISTIC-EQUATION-004 · ★☆☆ Basic
Solve \(y''+2y'+y=0\).
Choose one option.
Show answer
- D. \(y=(c_1+c_2t)e^{-t}\)
Explanation
\(r^2+2r+1=(r+1)^2=0\): double root \(r=-1\). The second independent solution is \(te^{-t}\).
Review the concept: Characteristic equation
The characteristic equation of \(ay''+by'+cy=0\), with constant coefficients \(a,b,c\), is the algebraic equation in the unknown \(r\) that results from trying \(y=e^{rt}\).
\[ar^2+br+c=0\]
More practice problems: characteristic equation
- ZW-CHARACTERISTIC-EQUATION-001 What is the characteristic equation of y''-5y'+6y=0?
- ZW-CHARACTERISTIC-EQUATION-002 Solve y''-y'-6y=0.
- ZW-CHARACTERISTIC-EQUATION-003 Solve y''+4y=0.
- ZW-CHARACTERISTIC-EQUATION-005 What are the roots of the characteristic equation of y''-4y=0?
- ZW-CHARACTERISTIC-EQUATION-006 Solve y''+2y'+5y=0.
- ZW-CHARACTERISTIC-EQUATION-007 An equation y''+by'+cy=0 has characteristic roots 1 and -4. Which equation is it?
- ZW-CHARACTERISTIC-EQUATION-008 For which values of b does y''+by'+9y=0 have a double characteristic root?
- ZW-CHARACTERISTIC-EQUATION-009 Solve y''-3y'+2y=0 with y(0)=1, y'(0)=0.
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