Solve no.4 inverse laplace L^{-1}left{s lnleft(frac{s}{sqrt{s^2+1}}right)+cot^{-1s}right}

Question
Laplace transform
asked 2021-02-25
Solve no.4 inverse laplace
\(L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1s}\right\}\)

Answers (1)

2021-02-26
Step 1
Given:
\(L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1s}\right\}\) Step 2
Use the linear property of inverse Laplace transformation, we get
\(L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1}s\right\}=L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\}+L^{-1}\left\{\cot^{-1}s\right\}\)
Now first find the inverse Laplace transformation of the first term
We know that the Laplace transformation formula:
\( \text{If } L\left\{f(t)\right\}=F(s) \text{ then } L\left\{t f(t)\right\}=\frac{-d}{ds}(F(s))\)
\(\Rightarrow f(t)=-\frac{1}{t}L^{-1}\left\{\frac{d}{ds}F(s)\right\}\)
\(L^{-1}\left\{s \ln \left(\frac{s}{\sqrt{s^2+1}}\right)\right\}=-\frac{1}{t}L^{-1}\left\{\frac{d}{s}\left(s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right)\right\}\)
Step 3
compute the derivative, we get
\(\Rightarrow -\frac{1}{t}L^{-1}{\ln\frac{s}{\sqrt{s^2+1}+\frac{1}{s^2+1}}}\)
\(\Rightarrow -\frac{1}{t}L^{-1}\left\{\ln\frac{s}{\sqrt{s^2+1}}\right\}-\frac{1}{t}L^{-1}\left\{\frac{1}{s^2+1}\right\}\)
Now we know that \(L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin(at)\)
\(\Rightarrow \frac{-1}{t}L^{-1}\left\{\ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\} -\frac{\sin(t)}{t}\)
Use the quotient rule of logarithmic, we get
\(\Rightarrow -\frac{1}{t}L^{-1}\left\{\ln(s)-\frac{1}{2}\ln(s^2+1)\right\} -\frac{\sin(t)}{t}\)
Step 4
Use the inverse Laplace transformation formula in the first term:
\(f(t)=-\frac{1}{t}L^{-1}\left\{\frac{d}{ds}F(s)\right\}\)
\(\Rightarrow \frac{1}{t^2}L^{-1}\left\{\frac{d}{ds}\left[\ln(s)-\frac{1}{2}ln(s^2+1)\right]\right\}-\frac{sin(t)}{t}\)
\(\Rightarrow \frac{1}{t^2}L^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{t^2}L^{-1}\left\{\frac{s}{s^2+1}\right\}-\frac{sin(t)}{t}\)
\(\Rightarrow \frac{1}{t^2}\left(L^{-1}\left\{{\frac{1}{s}}\right\}L^{-1}\left\{{\frac{s}{s^2+1}}\right\}\right)-\frac{\sin(t)}{t}\)
\(L^{-1}\left\{s \ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\}=\frac{1}{t^2}(1-\cos(t))-\frac{\sin(t)}{t}\)
Step 5
Now find the inverse Laplace transformation of \(\cot^{-1}(s)\)
We know that \(\cot^{-1}(s)=\tan^{-1}(\frac{1}{s})\)
\(\Rightarrow L^{-1}\left\{\cot^{-1}s\right\}=L^{-1}\left\{\tan^{-1}\left(\frac{1}{s}\right)\right\}\)
\(\Rightarrow L^{-1}\left\{\tan^{-1}\left(\frac{1}{s}\right)\right\}=-\frac{1}{t}L^{-1}\left\{\frac{d}{ds}\tan^{-1}\left(\frac{1}{s}\right)\right\}\)
\(=-\frac{1}{t}L^{-1}\left\{\frac{1}{1+\left(\frac{1}{s}\right)^2} \cdot\left(\frac{-1}{s^2}\right)\right\}\)
\(=-\frac{1}{t}L^{-1}\left\{\frac{-1}{s^2+1}\right\}\)
\(\Rightarrow L^{-1}\left\{\cot^{-1}s\right\}=\frac{(\sin t)}{t}\)
Step 6
Now add the inverse Laplace transformation of both terms
\(L^{-1}\left\{s\ln\left(\frac{s}{\sqrt{s^2+1}}\right)\right\}=\frac{1}{t^2}(1-\cos(t))-\frac{\sin(t)}{t}\)
\(L^{-1}\left\{\cot^{-1}s\right\}=\frac{\sin(t)}{t}\)
\(\Rightarrow L^{-1}\left\{s\ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1}s\right\}=\frac{1}{t^2}(1-\cos(t))-\frac{\sin(t)}{t}+\frac{\sin(t)}{t}\)
\(\text{Therefore, }\)
\(L^{-1}{s\ln\left(\frac{s}{\sqrt{s^2+1}}\right)+\cot^{-1}s}=\frac{(1-\cos t)}{t^2}\)
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