Find inverse Laplace transform of (e^2*e^(-4s))/(2s-1)

[jdoe]

I'm having trouble finding the inverse Laplace transform of $\displaystyle \frac{e^2e^{-4s}}{2s-1}$. The book suggests using the identity $\displaystyle \mathcal{L}^{-1}(F(ks))=\frac{1}{k}f(\frac{t}{k})$. Separating $\displaystyle e^2$ out, I get $\displaystyle e^2\mathcal{L}^{-1}(\frac{e^{-4s}}{2s-1})$. Letting $\displaystyle F(s)=\frac{e^{-2s}}{s-1}$, I get $\displaystyle e^2\mathcal{L}^{-1}(F(2s))$. Using the identity, the result should then be $\displaystyle \frac{e^2}{2}u_2(\frac{t}{2})e^{\frac{t}{2}}$. However, the book says the correct solution is $\displaystyle \frac{1}{2}e^{\frac{t}{2}}u_2(\frac{t}{2})$. The $\displaystyle e^2$ appears to have magically disappeared somehow.

This is problem number 29 section 6.3 from Elementary Differential Equations and Boundary Value Problems 10th edition by William E. Boyce and Richard C. DiPrima published by Wiley.

 

[Deleted member 4993]

I'm having trouble finding the inverse Laplace transform of (e^2*e^(-4s))/(2s-1). The book suggests using the identity L^(-1)(F(ks))=1/k*f(t/k). Seperating e^2 out, I get e^2*L^(-1)(e^(-4s)/(2s-1)). Letting F(s)=e^(-2s)/(s-1), I get e^2*L^(-1)(F(2s)). Using the identity, the result should then be e^2/2*u_2(t/2)*e^(t/2). However, the book says the correct solution is 1/2*e^(t/2)*u_2(t/2). The e^2 appears to have magically disappeared somehow.

This is problem number 29 section 6.3 from Elementary Differential Equations and Boundary Value Problems 10th edition. ...... Author? Publisher?

P.S. Does this forum support LaTeX?... Yes

.

 

[alain]

I'm having trouble finding the inverse Laplace transform of $\displaystyle \frac{e^2e^{-4s}}{2s-1}$. The book suggests using the identity $\displaystyle \mathcal{L}^{-1}(F(ks))=\frac{1}{k}f(\frac{t}{k})$. Separating $\displaystyle e^2$ out, I get $\displaystyle e^2\mathcal{L}^{-1}(\frac{e^{-4s}}{2s-1})$. Letting $\displaystyle F(s)=\frac{e^{-2s}}{s-1}$, I get $\displaystyle e^2\mathcal{L}^{-1}(F(2s))$. Using the identity, the result should then be $\displaystyle \frac{e^2}{2}u_2(\frac{t}{2})e^{\frac{t}{2}}$. However, the book says the correct solution is $\displaystyle \frac{1}{2}e^{\frac{t}{2}}u_2(\frac{t}{2})$. The $\displaystyle e^2$ appears to have magically disappeared somehow.

This is problem number 29 section 6.3 from Elementary Differential Equations and Boundary Value Problems 10th edition by William E. Boyce and Richard C. DiPrima published by Wiley.

You are looking at finding $\displaystyle e^2\mathcal{L}^{-1} \{\frac{e^{-4s}}{(2s-1)}\}$. I am not sure how using the identity you mention can get you to the answer without using the shifting theorem. The shifting theorem states that $\displaystyle \mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$, where u(t-a) is the unit step function such that u(t-a)=0 if $\displaystyle t < a$, and u(t-a)=1 if $\displaystyle t\ge a$. In $\displaystyle \mathcal{L}^{-1} \{\frac{e^{-4s}}{(2s-1)}\}$, a =4 and $\displaystyle F(s)=\frac{1}{(2s-1)}$. Then $\displaystyle f(t)= \mathcal{L}^{-1} \{F(s)\}=\frac{1}{2}e^{\frac{t}{2}}$, and $\displaystyle f(t-4)=\frac{1}{2}e^{\frac{(t-4)}{2}}=\frac{1}{2}e^{-2}e^{\frac{t}{2}}$. When multiplying f(t-4) by $\displaystyle e^2$ and $\displaystyle u(t-4)$, you end up with $\displaystyle \frac{1}{2}e^{\frac{t}{2}}u(t-4)$, which is the answer in your book assuming that $\displaystyle u_2(t/2)$ in the book's notation $\displaystyle =u(t-4)$ here.

 

[Sina]

Before seing my solution, I advise everyone to try to solve it first. If confronted with any problem, take a peek and try again.

 

[Deleted member 4993]

Before seing my solution, I advise everyone to try to solve it first. If confronted with any problem, take a peek and try again.
View attachment 23630

You are responding to a 5⁺ year old post....