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Chapter 22: Problem 3

Find the inverse Laplace transform of (a) \(\frac{1}{2(s+3)}\) (b) \(\frac{1}{2 s+3}\) (c) \(\frac{3 s}{s^{2}+1}\) (d) \(\frac{-6}{s^{2}+9}\) (e) \(\frac{s+2}{s^{2}+4}\)

Short Answer

Expert verified
Question: Find the inverse Laplace transforms of the following functions: (a) \(\frac{1}{2(s+3)}\) (b) \(\frac{1}{2s+3}\) (c) \(\frac{3s}{s^2+1}\) (d) \(\frac{-6}{s^2+9}\) (e) \(\frac{s+2}{s^2+4}\) Answer: (a) \(\frac{1}{2}e^{-3t}\) (b) \(\frac{1}{3}e^{-\frac{3}{2}t}\) (c) \(3\cos{t}\) (d) \(-\frac{2}{3}\sin{3t}\) (e) \(\frac{1}{4} \cos{2t}+\frac{1}{4}\sin{2t}\)

Step by step solution

01

(a) Find the inverse Laplace transform of \(\frac{1}{2(s+3)}\)

To find the inverse Laplace transform of \(\frac{1}{2(s+3)}\), we can use the Laplace transform table. The given function is the Laplace transform of the exponential function \(e^{-3t}\) scaled by \(\frac{1}{2}\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{1}{2(s+3)}\right)=\frac{1}{2}e^{-3t}$$
02

(b) Find the inverse Laplace transform of \(\frac{1}{2s+3}\)

To find the inverse Laplace transform of \(\frac{1}{2s+3}\), we can rewrite the function as: $$\frac{1}{2s+3} = \frac{1/3}{s+\frac{3}{2}}$$ Now, we recognize that this function is similar to the Laplace transform of the exponential function \(e^{-\frac{3}{2}t}\) scaled by \(\frac{1}{3}\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{1}{2s+3}\right)=\frac{1}{3}e^{-\frac{3}{2}t}$$
03

(c) Find the inverse Laplace transform of \(\frac{3s}{s^2+1}\)

To find the inverse Laplace transform of \(\frac{3s}{s^2+1}\), we compare this function with the Laplace transform table and find that this function is the Laplace transform of the cosine function scaled by \(3\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{3s}{s^2+1}\right)=3\cos{t}$$
04

(d) Find the inverse Laplace transform of \(\frac{-6}{s^2+9}\)

To find the inverse Laplace transform of \(\frac{-6}{s^2+9}\), we can rewrite the function as: $$\frac{-6}{s^2+9} = -\frac{2}{3}\frac{3}{s^2+3^2}$$ We recognize that this function is similar to the Laplace transform of the sine function scaled by \(-\frac{2}{3}\) and where \(a=3\). So the inverse Laplace transform is: $$L^{-1}\left(\frac{-6}{s^2+9}\right)=-\frac{2}{3}\sin{3t}$$
05

(e) Find the inverse Laplace transform of \(\frac{s+2}{s^2+4}\)

To find the inverse Laplace transform of \(\frac{s+2}{s^2+4}\), we first rewrite it as a sum: $$\frac{s+2}{s^2+4} = \frac{1}{4}\frac{4s}{s^2+4^2}+\left(\frac{1}{2}-\frac{1}{4}\right)\frac{1}{s^2+4^2}$$ Now, we identify both functions as Laplace transforms of the cosine function and sine function, respectively, and find the inverse Laplace transforms: $$L^{-1}\left(\frac{s+2}{s^2+4}\right)=\frac{1}{4} \cos{2t}+\frac{1}{4}\sin{2t}$$

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Most popular questions from this chapter

Given \(\mathcal{L}\\{x\\}=X, x(0)=2, x^{\prime}(0)=3\) find the Laplace transform of each of the following expressions: (a) \(x^{\prime}\) (b) \(x^{\prime \prime}\) (c) \(2 x^{\prime}-x\) (d) \(3 x^{\prime \prime}+2 x^{\prime}-x\) (e) \(-2 x^{\prime \prime}+x^{\prime}+3 x\)

The Laplace transform of \(x(t)\) is \(X(s)\), \(x(0)=2, x^{\prime}(0)=3, x^{\prime \prime}(0)=-1\). Find the Laplace transform of (a) \(2 x^{\prime}-3 x\) (b) \(x^{\prime \prime}-2 x^{\prime}+3 x\) (c) \(x^{m}\) (d) \(x^{\prime \prime \prime}+2 x^{\prime \prime}+3 x^{\prime}-4 x\) (e) \(2 x^{m \prime}-3 x^{n}-7 x^{\prime}+6 x\)

The first shift theorem states that if \(\mathcal{L}(f(t)\\}=F(s)\), then $$ \mathcal{L}\left\\{\mathrm{e}^{-a t} f(t)\right\\}=F(s+a) $$ where \(a\) is a constant. (a) From the definition of the Laplace transform show that $$ \mathcal{L}\left\\{\mathrm{e}^{-a t} f(t)\right\\}=\int_{0}^{\infty} \mathrm{e}^{-(s+a) t} f(t) \mathrm{d} t $$ and hence prove the first shift theorem. (b) Use Table \(1.1\) in Block 1 and the first shift theorem to find \(\mathcal{L}\left\\{u(t-3) \mathrm{e}^{-7 t}\right\\}\) where \(u(t)\) is the unit step function.

Find the inverse Laplace transform of (a) \(\frac{3}{s+2}\) (b) \(\frac{-2}{s-1}\) (c) \(\frac{2}{(s+2)^{3}}\) (d) \(\frac{1}{s}+\frac{1}{s+1}\) (e) \(\frac{1}{(s-3)^{2}}+\frac{1}{(s-3)^{3}}\)

Find the Laplace transform of each of the following expressions: (a) \(2 t \mathrm{e}^{2 t}\) (b) \(1-t^{3} \mathrm{e}^{3 t}\) (c) \(\mathrm{e}^{t}(1+\sin t)\) (d) \(\mathrm{e}^{-2 t}(\sin 3 t+2 \cos 3 t)\) (e) \(t(\cos 2 t-3 \sin 2 t)\)
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