S silencecloak May 2008 138 0 May 13, 2010 #1 I need some help getting started on this problem. \(\displaystyle y^\prime - \int\limits_{0}^{t}cos(t-\tau)y(\tau) d\tau = 1, y(0) = 2\) This problem scream Convolution to me, but the \(\displaystyle y(\tau)\) is throwing me off. I have considered recursive integration, but am still not sure what to do with that term. Any help? Thanks!(Happy)

I need some help getting started on this problem. \(\displaystyle y^\prime - \int\limits_{0}^{t}cos(t-\tau)y(\tau) d\tau = 1, y(0) = 2\) This problem scream Convolution to me, but the \(\displaystyle y(\tau)\) is throwing me off. I have considered recursive integration, but am still not sure what to do with that term. Any help? Thanks!(Happy)

T TwistedOne151 Dec 2007 276 143 Anchorage, AK May 13, 2010 #2 Laplace Considering the form of the convolution, consider taking the Laplace transform of the equation. --Kevin C.

Laplace Considering the form of the convolution, consider taking the Laplace transform of the equation. --Kevin C.

Random Variable May 2009 959 362 May 13, 2010 #3 Using TwistedOne151's advice, let \(\displaystyle Y(s) = \mathcal{L} [y(t)] \) then \(\displaystyle sY(s) - y(0) - \mathcal{L} [\cos t ] \mathcal{L} [y(t)] = \frac{1}{s} \) \(\displaystyle sY(s) -2 - \frac{s}{s^{2}+1}Y(s) = \frac{1}{s} \) \(\displaystyle Y(s)\Big(\frac{s^{3}}{s^{2}+1} \Big) = \frac{1}{s} + 2 \) \(\displaystyle Y(s) = \frac{s^{2}+1}{s^{4}} + 2 \ \frac{s^{2}+1}{s^{3}} = \frac{1}{s^{2}} + \frac{1}{s^{4}} + \frac{2}{s} + \frac{2}{s^{3}}\) then \(\displaystyle y(t) = t + \frac{t^{3}}{6} + 2 + t^{2} \) Reactions: silencecloak

Using TwistedOne151's advice, let \(\displaystyle Y(s) = \mathcal{L} [y(t)] \) then \(\displaystyle sY(s) - y(0) - \mathcal{L} [\cos t ] \mathcal{L} [y(t)] = \frac{1}{s} \) \(\displaystyle sY(s) -2 - \frac{s}{s^{2}+1}Y(s) = \frac{1}{s} \) \(\displaystyle Y(s)\Big(\frac{s^{3}}{s^{2}+1} \Big) = \frac{1}{s} + 2 \) \(\displaystyle Y(s) = \frac{s^{2}+1}{s^{4}} + 2 \ \frac{s^{2}+1}{s^{3}} = \frac{1}{s^{2}} + \frac{1}{s^{4}} + \frac{2}{s} + \frac{2}{s^{3}}\) then \(\displaystyle y(t) = t + \frac{t^{3}}{6} + 2 + t^{2} \)

S silencecloak May 2008 138 0 May 13, 2010 #4 Thanks that was a lot easier than I thought. Makes perfect sense.