Question : Find Inverse Laplace transform of Y(s)=1/((s^2+7s+9))
Solution :
To Find Inverse Laplace transform of Y(s)=1/((s^2+7s+9)) Given function \[Y\left(s\right)=\frac{1}{(s^2+7s+9)}\] It can be written as \[Y\left(s\right)=\frac{1}{\left(s+\frac{7}{2}\right)^2-\frac{13}{4}}\] \[Y\left(s\right)=\frac{1}{\left(s+\frac{7}{2}\right)^2-\left(\sqrt{\frac{13}{2}}\right)^2}\] \[Y\left(s\right)=\frac{1}{\left(s+\frac{7}{2}-\sqrt{\frac{13}{2}}\right)\left(s+\frac{7}{2}+\sqrt{\frac{13}{2}}\right)}\] So the Partial fraction of Y(s) is given as \[Y\left(s\right)=\frac{1}{\sqrt{13}}\left[\frac{1}{s+\frac{7}{2}-\frac{\sqrt{13}}{2}}-\frac{1}{s+\frac{7}{2}+\frac{\sqrt{13}}{2}}\right]\] Applying inverse Laplace transeform \[L^{-1}\left(Y\left(s\right)\right)=\frac{1}{\sqrt{13}}L^{-1}\left[\frac{1}{s+\frac{7}{2}-\frac{\sqrt{13}}{2}}-\frac{1}{s+\frac{7}{2}+\frac{\sqrt{13}}{2}}\right]\] \[L^{-1}\left(Y\left(s\right)\right)=\frac{1}{\sqrt{13}}L^{-1}\left[\frac{1}{s+\frac{7}{2}-\frac{\sqrt{13}}{2}}\right]-\frac{1}{\sqrt{13}}L^{-1}\left[\frac{1}{s+\frac{7}{2}+\frac{\sqrt{13}}{2}}\right]\] We know that inverse Laplace transeform formulas \[L^{-1}\left[\frac{1}{s-a}\right]=e^{at}\ \ \ ,\ L^{-1}\left[\frac{1}{s+a}\right]=e^{-at}\ \] Therefore \[L^{-1}\left(Y\left(s\right)\right)=y\left(t\right)=\frac{1}{\sqrt{13}}\ \left[e^{\left(-\frac{7}{2}+\frac{\sqrt{13}}{2}\right)t}-e^{\left(-\frac{7}{2}-\frac{\sqrt{13}}{2}\right)t}\right]\] Is the required Inverse Laplace transform of Y(s)=1/((s^2+7s+9))