Question

find inverse laplace of 21-s/(s-5)(s+30

Answer 1

make partial fractions

Answer 2

Use partial fraction decomposition to rewrite it as the sum of nicer terms.$$\frac{s-21}{(s-5)(s+3)}=\frac{A}{s-5}+\frac{B}{s+3}\\A(s+3)+B(s-5)=s-21\\(A+B)s+3A-5B=s-21\\\quad\implies A+B=1,3A-5B=-21$$Since \(A+B=1\) we know \(B=1-A\):$$3A-5(1-A)=-21\\3A-5+5A=-21\\8A-5=-21\\8A=-16\\A=-2$$... and thus \(B=1-(-2)=3\).$$\frac{s-21}{(s-5)(s+3)}=-\frac2{s-5}+\frac3{s+3}$$Because our original expression was \(\dfrac{21-s}{(s-5)(s+3)}=-\dfrac{s-21}{(s-5)(s+3)}\) flip the signs:$$\frac{21-s}{(s-5)(s+3)}=\frac2{s-5}-\frac3{s+3}$$Now we can apply the inverse transform:$$\mathcal{L}^{-1}\left\{\frac2{s-5}-\frac3{s+3}\right\}=2\mathcal{L}^{-1}\left\{\frac1{s-5}\right\}-3\mathcal{L}^{-1}\left\{\frac1{s+3}\right\}$$Recall that $$\mathcal{L}\left\{e^{-at}\right\}=\frac1{s+a}\Longleftrightarrow\mathcal{L}^{-1}\left\{\frac1{s+a}\right\}=e^{-at}$$so we can finish easily now:$$2e^{5t}-3e^{-3t}$$

Answer 3

thank you :) i'll try to understand the solution :)