# Find the inverse Laplace Transform from the function below. F(s)=frac(7)((s+11)(s+12))

Find the inverse Laplace Transform from the function below
$$\displaystyle{F}{\left({s}\right)}={\frac{{{7}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}$$

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Fatema Sutton

Step 1
We have to find inverse laplace transformation of given function.
Step 2
Now,
$$\displaystyle{F}{\left({s}\right)}={\frac{{{7}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}$$
Now,
$$\displaystyle{F}{\left({s}\right)}={7}{\left({\frac{{{1}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}\right)}$$
$$\displaystyle{F}{\left({s}\right)}={7}{\left({\frac{{{1}}}{{{s}+{11}}}}-{\frac{{{1}}}{{{s}+{12}}}}\right)}$$
Therefore,
$$\displaystyle{L}^{{-{1}}}{F}={L}^{{-{1}}}{\left[{\frac{{{7}}}{{{s}+{11}}}}-{\frac{{{7}}}{{{s}+{12}}}}\right]}$$
$$\displaystyle={L}^{{-{1}}}{\left[{\frac{{{7}}}{{{s}+{11}}}}\right]}-{L}^{{-{1}}}{\left[{\frac{{{7}}}{{{s}+{12}}}}\right]}$$
$$\displaystyle={7}{e}^{{-{11}{t}}}-{7}{e}^{{-{12}{t}}}$$
Hence,
$$\displaystyle{L}^{{-{1}}}{\left({\frac{{{7}}}{{{\left({s}+{11}\right)}{\left({s}+{12}\right)}}}}\right)}={7}{e}^{{-{11}{t}}}-{7}{e}^{{-{12}{t}}}$$