Monday, 22 April 2019

4:31 PM

Gives us a pair of complex conjugate roots, so we know to use this method

𝑋﷐𝑠﷯=﷐10﷐𝑠﷮2﷯+15𝑠−5﷮𝑠﷐﷐𝑠﷮2﷯+2𝑠+5﷯﷯=﷐𝐴﷮𝑠﷯+﷐𝐵𝑠+𝐶﷮﷐𝑠﷮2﷯+2𝑠+5﷯ 𝐴﷐﷐𝑠﷮2﷯+2𝑠+5﷯+﷐𝐵𝑠+𝐶﷯𝑠=10﷐𝑠﷮2﷯+15𝑠−5 5𝐴=−5;𝐴=−1 ﷐𝐵𝑠+𝐶﷯𝑠−﷐𝑠﷮2﷯−2𝑠−5=10﷐𝑠﷮2﷯+15𝑠−5 𝐵=11;𝐶=17 So 𝑋﷐𝑠﷯=−﷐1﷮𝑠﷯+﷐11𝑠+17﷮﷐𝑠﷮2﷯+2𝑠+5﷯ Complete the square so we can get an identity 𝑋﷐𝑠﷯=−﷐1﷮𝑠﷯+﷐11﷐𝑠+1﷯+6﷮﷐﷐𝑠+1﷯﷮2﷯+4﷯ 𝑋﷐𝑠﷯=−﷐1﷮𝑠﷯+11﷐﷐𝑠+1﷯﷮﷐﷐𝑠+1﷯﷮2﷯+﷐2﷮2﷯﷯+3﷐2﷮﷐﷐𝑠+1﷯﷮2﷯+﷐2﷮2﷯﷯ 𝑥﷐𝑡﷯=﷐−1+11﷐𝑒﷮−𝑡﷯﷐cos﷮2𝑡﷯+3﷐𝑒﷮−𝑡﷯﷐sin﷮2𝑡﷯﷯𝑢(𝑡) Ink Drawings Ink Drawings

 

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