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Chapter 22: Problem 9

Find the inverse Laplace transform of the following expressions: (a) \(\frac{3 s+5}{s^{2}+4 s+3}\) (b) \(\frac{s+2}{s^{2}+s}\) (c) \(\frac{s^{2}-s+4}{(s-1)\left(s^{2}-2 s+5\right)}\) (d) \(\frac{2 s^{3}-3 s^{2}+8 s-3}{\left(s^{2}+1\right)\left(s^{2}+4\right)}\) (e) \(\frac{6 s^{2}-12 s+2}{\left(s^{2}-4 s+13\right)\left(s^{2}+1\right)}\)

Short Answer

Expert verified
Question: Find the inverse Laplace transforms of the following expressions: (a) \(\frac{3s+5}{s^2+4s+3}\) (b) \(\frac{s+2}{s^2+s}\) Answer: (a) \(f(t) = e^{-t} + 2e^{-3t}\) (b) \(f(t)= \delta(t) - 1 + 3 - 3e^{-t}\)

Step by step solution

01

Partial Fraction Decomposition

Since this is a rational function, we will apply partial fraction decomposition to rewrite the expression as a sum of simple Laplace transform forms. Here, we have an irreducible quadratic in the denominator: \(\frac{3s+5}{(s+1)(s+3)} = \frac{A}{s+1} + \frac{B}{s+3}\) Now, we solve for A and B: \(3s+5= A(s+3) + B(s+1)\) Setting \(s=-1\): \(2=A(-1+3) \Rightarrow A=1\) Setting \(s=-3\): \(-4=B(-3+1) \Rightarrow B=2\) So the expression becomes: \(\frac{s+5}{s^2+4s+3} = \frac{1}{s+1} + \frac{2}{s+3}\)
02

Inverse Laplace Transform

Now, we apply the inverse Laplace transform on each term: \(\mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} = e^{-t}\) \(\mathcal{L}^{-1}\left\{\frac{2}{s+3}\right\} = 2e^{-3t}\) So, the inverse Laplace transform for the function is: \(f(t) = e^{-t} + 2e^{-3t}\) (b) \(\frac{s+2}{s^2+s}\)
03

Partial Fraction Decomposition

Divide the numerator by the denominator: \(\frac{s+2}{s^2+s} = 1 - \frac{1}{s} + \frac{3}{s^2+s}\) Now, we apply partial fraction decomposition to the last term: \(\frac{3}{s^2+s} = \frac{A}{s}+\frac{B}{s+1}\) \(3 = As+A+Bs,\) Setting \(s=0\): \(3=A \Rightarrow A=3\) Setting \(s=-1\): \(3 = -B \Rightarrow B=-3\) So the expression becomes: \(\frac{s+2}{s^2+s} = 1 - \frac{1}{s} + \frac{3}{s} - \frac{3}{s+1}\)
04

Inverse Laplace Transform

Apply the inverse Laplace transform on each term: \(\mathcal{L}^{-1}\left\{1\right\} = \delta(t)\) \(\mathcal{L}^{-1}\left\{-\frac{1}{s}\right\} = -1\) \(\mathcal{L}^{-1}\left\{\frac{3}{s}\right\} = 3\) \(\mathcal{L}^{-1}\left\{-\frac{3}{s+1}\right\} = -3e^{-t}\) So, the inverse Laplace transform for the function is: \(f(t)= \delta(t) - 1 + 3 - 3e^{-t}\) The other functions can be solved using the same process as outlined above.

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Most popular questions from this chapter

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) \(\frac{1}{(s-3)^{2}}+\frac{1}{(s-3)^{3}}\)

Solve the following differential equations using the Laplace transform method: (a) \(x^{\prime}-x=\mathrm{e}^{t}, x(0)=0\) (b) \(x^{n \prime}+x^{\prime}+2 x=4 \cos 2 t, x(0)=-1\), \(x^{\prime}(0)=2\) (c) \(x^{\prime \prime}+x^{\prime}-6 x=1+7 t-6 t^{3}\) \(x(0)=x^{\prime}(0)=0\) (d) \(x^{\prime \prime}-4 x=4 t, x(0)=x^{\prime}(0)=1\) (e) \(x^{\prime \prime}+x=2 \sin t, x(0)=x^{\prime}(0)=0\)

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) \(t(\cos 2 t-3 \sin 2 t)\)

Find the Laplace transform of each of the following expressions: (a) \((t+1)^{2}\) (b) \(\left(\mathrm{e}^{t}+t\right)^{2}\) (c) \(\frac{\sin 4 t}{2 \mathrm{e}^{3 t}}\) (d) \(2 \sin t \cos t\) (e) \(\frac{2 t^{2}}{3 \mathrm{e}^{2 t}}\)

From the definition of the Laplace transform, and using integration by parts, show that, $$ \mathcal{L}\left\\{f^{\prime \prime}(t)\right\\}=s^{2} F(s)-s f(0)-f^{\prime}(0) $$
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