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Soln:- \mathbb{H}^{-}\left\{\frac{4 s+10}{s^{2}+5 s+6} e^{-2 s}\right\}first we take the partial fraction we get.\begin{array}{l}\frac{4 s+10}{s^{2}+5 s+6}=\frac{4 s+10}{(s+2)(s+3)} \\\frac{4 s+10}{(s+2)(s+3)}=\frac{A}{(s+2)}+\frac{B}{(s+3)} \\4 s+10=A(s+3)+B(s+2) . \\4 s+10=(A+B) s+(3 A+2 B) \\A+B=4, \quad S A+2 B=10\end{array}Solving this we get A=2 and B=2.\begin{array}{l} \frac{4 s+10}{(s+2)(s+3)}=\frac{2}{s+2}+\frac{2}{s+3} \\\therefore\left\{\frac{2}{s+2}+\frac{2}{s+3}\right\}\end{array}lsing linear property we get.\begin{array}{l}\text { L. }\left\{\frac{s s+10}{(s+2)(s+3)}\right\}=2 L\left\{\left\{\frac{1}{s+2}\right\}+2 t+\left\{\frac{1}{s+3}\right\}\right. \\=2 e^{-2 t}+2 e^{-3 t} \text {. (1) } \\\end{array}Apply inverse transform fule: if (1\{F(1))=f(t) then1 *\left\{e^{-a s} f(s)\right\}=h(t-a) f(t-a)where H(t) is heaviside step funcsion.=H(t-2) t^{-1}\left\{\frac{s+10}{s^{2}+51+6}\right\}(t-2) \text {. }hene laplace inverse transform in u(t-c) of e^{-x} \frac{\$ s+10}{s^{2}+5 s+6} is=v(t-2) 2 e^{-2(t-2)}+u(t-2) e^{-3(t-2)}=2 U(t-2)\left(e^{-2(t-2)}+e^{-3(t-2)}\right) \text {. Req sor.. }if you found result is usefure then mz like ...