text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by } F(s)=int_0^infty e^{-st} f(t)dt text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} text{ , the following improper integral is evaluated using integration by parts:} F(s)=int_0^infty e^{-st}e^{-t}dt=int_0^infty e^{-(s+1)t}dt=frac{1}{s+1} text{ Verify the following Laplace transforms, where u is a real number. } f(t)=t rightarrow F(s)=frac{1}{s^2}

text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by } F(s)=int_0^infty e^{-st} f(t)dt text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} text{ , the following improper integral is evaluated using integration by parts:} F(s)=int_0^infty e^{-st}e^{-t}dt=int_0^infty e^{-(s+1)t}dt=frac{1}{s+1} text{ Verify the following Laplace transforms, where u is a real number. } f(t)=t rightarrow F(s)=frac{1}{s^2}

Question
Laplace transform
asked 2021-03-02
\(\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }\)
\(F(s)=\int_0^\infty e^{-st} f(t)dt
\(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:}
\(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}\)
\(\text{ Verify the following Laplace transforms, where u is a real number. }\)
\(f(t)=t \rightarrow F(s)=\frac{1}{s^2}\)

Answers (1)

2021-03-03
\(\text{Step 1}\)
\(\text{Here, the objective is to prove that }\ L(t)=\frac{1}{s^2}\)
\(\text{Step 2}\)
\(\text{Let } f(t) \text{ be t}\)
\(\text{Laplace transformation of f(t) is defined as:}\)
\(L(t)=F(s)=\int_0^\infty e^{-st}f(t)dt\)
\(\text{Substitute } f(t)=t\)
\(L(t)=F(s)=\int_0^\infty e^{-st} tdt\)
\(\text{Use integration by part } \int fg'=fg-\int f'g\)
\(\text{Here } f=t \Rightarrow f'=1 \text{ and } g'=e^{-st}\Rightarrow g=-\frac{e^{-st}}{s}\)
\(F(s)=\left[-\frac{e^{-st}t}{s}\right]_0^\infty -\int_0^\infty -\frac{e^{-st}}{s}dt \ (Where \int e^x dx=e^x+c)\)
\(F(s)=0+\int_0^\infty \frac{e^{-st}}{s}dt\)
\(F(s)=\left[-\frac{e^{-st}}{s^2}\right]_0^\infty\)
\(F(s)=\left[\(\lim_{t\rightarrow\infty}-\frac{e^{-st}}{s^2}-\bigg(-\frac{e^0}{s^2}\bigg)\right]\)
\(F(s)=\left[0-\bigg(-\frac{1}{s^2}\bigg)\right]\)
\(\text{Hence }\)
\(f(t)=t \rightarrow F(s)=\frac{1}{s^2}\)
0

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