Exercise · ZW-CONVOLUTION-010 · ★★★ Advanced
The solution of \(y''+y=g(t)\), \(y(0)=y'(0)=0\), is \(y=\sin t*g\). For \(g=1\), what is \(y(\pi)\)?
Choose one option.
Show answer
- A. \(2\)
Explanation
With \(g=1\), \(y=1-\cos t\) (earlier exercise), so \(y(\pi)=1-(-1)=2\). In the transform, \(Y=\dfrac{G}{s^2+1}\).
Review the concept: Convolution
The convolution \(f*g\) of two functions \(f\) and \(g\), defined for \(t\ge0\), is the function of \(t\) that integrates from \(0\) to \(t\) the product of \(f(\tau)\) and a reflected, shifted copy of \(g\).
\[(f*g)(t)=\int_0^t f(\tau)\,g(t-\tau)\,d\tau\]
More practice problems: convolution
- ZW-CONVOLUTION-001 What is (1*1)(t)?
- ZW-CONVOLUTION-002 What is (t*1)(t)?
- ZW-CONVOLUTION-003 What is (e^t*1)(t)?
- ZW-CONVOLUTION-004 If F=mathcal L\f\ and G=mathcal L\g\, what is mathcal L\f*g\?
- ZW-CONVOLUTION-005 Use the convolution theorem: what is mathcal L\1*t\?
- ZW-CONVOLUTION-006 What is mathcal L^-1\(1)/(s(s+1))\, viewed as 1*e^-t?
- ZW-CONVOLUTION-007 What is mathcal L^-1\(1)/(s^2(s-1))\?
- ZW-CONVOLUTION-008 What is (sin t*1)(t)?