Laplace Transform – 1 Exercise 1
1. Define Integral transform and Laplace transform.
Solution:
Integral Transform: Let K(p, t) be a function of two variables p and t, where p is a parameter real or complex independent of t. The function f(p) defined by the integral \(\int_{-\infty}^{\infty} K(p, t) F(t) d t\) is called the integral transform of the function F(t) and is denoted by T[F(t)].
The function K(p, t) is called the kernel of the transformation.
Laplace transform: Let a function F(t) be continuous and defined for all positive values of t. The Laplace transformation or Laplace transform of F(t) is defined by \(L[F(t)]=\int_0^{\infty} e^{-p t}. F(t) d t\) where ‘ p’ is a parameter, and the transform is denoted by f(p) or \(\bar{F}(p)\).
2. Define piecewise continuous function.
Solution:
Piecewise continuous function
A function F(t) is said to be piecewise continuous or sectionally continuous in a closed interval [a, b] if it is defined in that interval and is such that the interval can be divided into a finite number of sub-intervals in each of which F(t) is continuous and has finite left and right-hand limits.
3. Define a function of exponential order a.
Solution:
A function of exponential order a
A function F(t) is said to be of exponential order a as t→∞ if there exists a positive real number M, a number a, and a finite number \(t_0\) such that \(\left|e^{-a t} F(t)\right|<M\) i. e., \(|F(t)|<M e^{a t} \forall t \geq t_0\).
4. Define a function of class A.
Solution:
A function of class A
A function F(t) is said to be of class A if F(t) is piecewise continuous on every finite interval in the range t≥0 and is of exponential order as t→∞
5. If a function F(t) is piecewise continuous on every finite interval in the range t≥0 and satisfies \(|F(t)|<M e^{a t} \forall t \geq t_0\) and for some constants a and M, then prove that the Laplace transform of F(t) exists for all p>a.
Solution:
Given
If a function F(t) is piecewise continuous on every finite interval in the range t≥0 and satisfies \(|F(t)|<M e^{a t} \forall t \geq t_0\) and for some constants a and M,
L[F(t)] = \(\int_0^{\infty} e^{-p t} F(t) d t=\int_0^{t_0} e^{-p t} F(t) d t+\int_{t_0}^{\infty} e^{-p t} F(t) d t \rightarrow \text { (1) }\)
Since F(t) is piecewise continuous on \(\left[0, t_0\right]\), the integral \(\int_0^{t_0} e^{-p t} F(t) d t\) exists.
∴ \(\int_{t_0}^{\infty} e^{-p t} F(t) d t|\leq \int_{t_0}^{\infty}| e^{-p t} F(t) d t\)l,< \(\int_{t_0}^{\infty} e^{-p t} M e^{a t} d t\)
= \(\int_{t_0}^{\infty} M e^{-(p-a) t} d t=M [\frac{e^{-(p-a) t}}{-(p-a)}]_{t_0}^{\infty}\)
= \(\frac{M e^{-(p-a) t_0}}{p-a}\) for p>a .
But \(\frac{M e^{-(p-a) t_0}}{p-a}\) can be made as small as we require by taking \(t_0\) sufficiently large. Thus L[F(t)] exists for all p>a.
6. Show that \(L\left(\frac{1}{\sqrt{t}}\right)\) exists for p≥0; even if \(\frac{1}{\sqrt{t}}\)t is not piecewise continuous in the range t≥0. Does this function belong to a function of class A? Justify your answer.
Solution:
The function \(F(t)=\frac{1}{\sqrt{t}}\) is not piecewise continuous on every finite interval in the range t≥0. But
L[F(t)] = \(\int_0^{\infty} e^{-p t} F(t) d t=\int_0^{\infty} e^{-p t} \frac{1}{\sqrt{t}} d t=\int_0^{\infty} e^{-x^2} d x\) where \(\sqrt{p t}=x\)
= \(\frac{2}{\sqrt{p}} \frac{\sqrt{\pi}}{2}=\frac{\sqrt{\pi}}{\sqrt{p}}, p>0\) ∴L[F(t)] exists for p>0.
7. If the Laplace transforms of two functions \(F_1(t), F_2(t)\)) exists then prove that \(L\left[F_1(t)+F_2(t)\right]=L\left[F_1(t)\right]+L\left[F_2(t)\right]\).
Solution:
L\(\left[F_1(t)+F_2(t)\right]=\int_0^{\infty}\left[F_1(t)+F_2(t)\right] e^{-p t} \cdot d t\)
= \(\int_0^{\infty} F_1(t) e^{-p t} d t+\int_0^{\infty} F_2(t) e^{-p t} d t\)
= \(L\left[F_1(t)\right]+L\left[F_2(t)\right]\)
8. If \(k_1, k_2\) are two constants and the Laplace transforms of the functions \(F_1(t), F_2(t)\) exists then prove that \(L\left[k_1 F_1(t)+k_2 F_2(t)\right]=k_1 L\left[F_1(t)\right]+k_2 L\left[F_2(t)\right]\).
Solution:
Given
If \(k_1, k_2\) are two constants and the Laplace transforms of the functions \(F_1(t), F_2(t)\)
L\(\left[k_1 F_1(t)+k_2 F_2(t)\right]=\int_0^{\infty}\left[k_1 F_1(t)+k_2 F_2(t)\right] e^{-p t} \cdot d t\)
= \(\int_0^{\infty} k_1 F_1(t) e^{-p t} d t+\int_0^{\infty} k_2 F_2(t) e^{-p t} d t\)
= \(k_1 \int_0^{\infty} F_1(t) e^{-p t} d t+k_2 \int_0^{\infty} F_2(t) e^{-p t} d t\)
= \(k_{\mathrm{i}} L\left[F_1(t)\right]+k_2 L\left[F_2(t)\right]\)
9. Show that the Laplace transform of the unit function F(t)=1 is 1/p.
Solution:
L[F(t)] = \(\int_0^{\infty} 1 \cdot e^{-p t} d t=\left[\frac{e^{-p t}}{-p}\right]_0^{\infty}=\frac{1}{p}, \text { provided } p>0\)
10. Show that Laplace transforms of \(F(t)=t^n,-1<n<0\) exists although it is not a function of class A.
Solution:
Here F(t) \(\rightarrow \infty\) as \(t \rightarrow 0\) for \(t \geq 0\), i.e. the function is not piecewise continuous on every finite interval in the range \(t \geq 0\).
We have \(\underset{t \rightarrow \infty}{\text{Lt}}\left\{e^{a t} F(t)\right\}=\underset{t \rightarrow \infty}{\text{Lt}}\left(\frac{t^n}{e^{a t}}\right)=\underset{t \rightarrow \infty}{\text{Lt}} \frac{1}{t^n e^{a t}}=\underset{t \rightarrow \infty}{L t} \frac{1}{t^m e^{a t}}\) where 0<m<1 =0, if a>0.
∴ F(t) = \(t^n\) is of exponential order.
Since \(F(t)=t^n\) is not piecewise continuous over every finite interval in the range \(t \geq 0\), hence it is not a function of class A. But \(t^n\) is integrable from o to any positive number \(t_0\).
Now \(L\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^{\infty} e^{-p t} t^n d t=\int_0^{\infty} e^{-x}\left(\frac{x}{p}\right)^n \frac{1}{p} d x\)
putting pt=x so that \(d t=\frac{1}{p} d x\) and taking p>0
= \(\frac{1}{p^{n+1}} \int_0^{\infty} e^{-x} x^{(n+1)-1} d x=\frac{\Gamma(n+1)}{p^{n+1}}\), if p>0 and n+1>0 i.e. n>-1.
Hence the Laplace transform of \(t^n, 0>n>-1\) exists, although it is not a function of class A.
11. If n is a positive integer, then show that \(L\left(t^n\right)=\frac{n !}{p^{n+1}}, p>0\)
Solution:
Given
n is a positive integer
L\(\left(t^n\right)=\int_0^{\infty} t^n e^{-p t} d t=\left[\frac{t^n \cdot e^{-p t}}{-p}\right]_0^{\infty}-\int_0^{\infty} \frac{e^{-p t}}{-p} \cdot n t^{n-1} d t=\frac{n}{p} \int_0^{\infty} e^{-p t} t^{n-1} d t=\frac{n}{p} L\left(t^{n-1}\right)\)
Similarly \(L\left(t^{n-1}\right)=\frac{n-1}{p} \cdot L\left(t^{n-2}\right)\) and so on.
Hence \(L\left(t^n\right)=\frac{n}{p} \cdot \frac{n-1}{p} \cdot \frac{n-2}{p} \cdots \frac{2}{p} \cdot \frac{1}{p} \cdot L\left(t^{n-n}\right)=\frac{n !}{p^n} L\left(t^{\circ}\right)=\frac{n !}{p^n} L(1)=\frac{n !}{p^n} \cdot \frac{1}{p}=\frac{n !}{p^{n+1}}\)
12. If ‘ a ‘ is a constant, then show that \(L\left(e^{a t}\right)=\frac{1}{p-a}\)
Solution:
L\(\left(e^{a t}\right)=\int_0^{\infty} e^{-p t} e^{a t} d t=\int_0^{\infty} e^{-(p-a) t} d t=-\left[\left.\frac{e^{-(p-a) t}}{(p-a)}\right|_0 ^{\infty}=\frac{1}{p-a}, \text { provided } p>a\right.\)
13. Find the Laplace transform of \(e^{-a t}\) by using the basic definition.
Solution:
L\(\left(e^{-a t}\right)=\int_0^{\infty} e^{-p t} e^{-a t} d t=\int_0^{\infty} e^{-(p+a) t} d t=-\left[\left.\frac{e^{-(p+a) t}}{(p+a)}\right|_0 ^{\infty}=\frac{1}{p+a}, \text { provided } p>a\right.\)
14. If ‘ a ‘ is a constant, then show that \(L(\sin a t)=\frac{a}{p^2+a^2}\).
Solution:
L\((\sin a t)=\int_0^{\infty} e^{-p t} \sin a t d t=\left[\frac{e^{-p t}}{p^2+a^2}(-p \sin a t-a \cos a t)\right]_0^{\infty}=\frac{a}{p^2+a^2}\)
15. If ‘ a ‘ is a constant, then show that \(L(\cos a t)=\frac{p}{p^2+a^2}\).
Solution:
L\((\cos a t)=\int_0^{\infty} e^{-p x} \cos a t d t=\left[\frac{e^{-p t}}{p^2+a^2}(-p \cos a t+a \sin a t)\right]_0^{\infty}=\frac{p}{p^2+a^2}\)
16. If ‘ a ‘ is a constant, then show that \(L(\sinh a t)=\frac{a}{p^2-a^2}\).
Solution:
L\((\sinh a t)=\int_0^{\infty} e^{-\not t} \sinh a t d t=\int_0^{\infty} e^{-p t}\left[\frac{e^{a t}-e^{-a t}}{2}\right] d t=\frac{1}{2}\left[\int_0^{\infty} e^{-p t} e^{a t} d t-\int_0^{\infty} e^{-p t} e^{-a t} d t\right]\)
= \(\frac{1}{2}\left[\frac{1}{p-a}-\frac{1}{p+a}\right]=\frac{a}{p^2-a^2}\)
17. If ‘ a ‘ is a constant, then show that \(L(\cosh a t)=\frac{p}{p^2-a^2}\).
Solution:
L\((\cosh a t)=\int_0^{\infty} e^{-p t} \cosh a t d t=\int_0^{\infty} e^{-p t}\left[\frac{e^{a t}+e^{-a t}}{2}\right] d t=\frac{1}{2}\left[\int_0^{\infty} e^{-p t} e^{a t} d t+\int_0^{\infty} e^{-p t} e^{-a t} d t\right]\)
= \(\frac{1}{2}\left[\frac{1}{p-a}+\frac{1}{p+a}\right]=\frac{p}{p^2-a^2}\)
18. Define a gamma function.
Solution:
Gamma function
If n>0 then the Gamma function Γ(n) is defined as \(\Gamma(n)=\int_0^{\infty} e^{-x} x^{n-1} d x\).
19. Define Heaviside’s unit step function.
Solution:
Heaviside’s unit step function
The Heavisides unit step function is denoted by U(t-a) or u(t-a) or H(t-a) and is defined as H(t-a)=0 if t<a and H(t-a)=1 if t>a.
20. Show that \(L[H(t-a)]=\frac{e^{-a p}}{p}\).
Solution:
L\(\left[H(t-a)=\int_0^{\infty} e^{-p t} H(t-a) d t=\int_0^a e^{-p t} H(t-a) d t+\int_a^{\infty} e^{-p t} H(t-a) d t\right.\)
= \(\int_0^a e^{-p t}(0) d t+\int_a^{\infty} e^{-p t} 1 d t=\int_a^{\infty} e^{-p t} d t=\left[\frac{e^{-p t}}{-p}\right]_a^{\infty}=\frac{e^{-a p}}{p} .\)
21. Define Unit Impulse Function.
Solution:
Unit Impulse Function
The Unit Impulse Function or Dirac Delta Function at a is denoted by δ(t-a) and is defined as \(\delta(t-a)=\underset{h \rightarrow 0}{L t} I(h, t-a)\), where \(I(h, t-a)=\frac{1}{h}\) if a<t<a+h; I(h,t-a)=0 for t<a or t>a+h.
22. Show that \(L[\delta(t-a)]=e^{-a p}\).
Solution:
We know that I(h, t-a) = \(\frac{1}{h}[H(t-a)-H(t-a-h)]\)
L\([I(h, t-a)]=\frac{1}{h}\{L[H(t-a)]-L[H(t-a-h)]\}=\frac{1}{h}\left[\frac{e^{-a p}}{p}-\frac{e^{-(a+h) p}}{p}\right]=\frac{e^{-a p}}{p}\left[1-e^{-h p}\right]\)
L\(\left[\delta(t-a)=\underset{h \rightarrow 0}{L t} L[I(h, t-a)]={ }_{h \rightarrow 0}^{L t} \frac{e^{-a p}}{p h}\left[1-e^{-h p}\right]=e^{-a p}\right.\)
23. Prove that the function \(F(t)=t^2\) is of exponential order 3 .
Solution:
⇒ \(\ Lim_{t \rightarrow \infty} e^{-3 t} F(t)=\ Lim_{t \rightarrow \infty} e^{-3 t} \cdot t^2=\ Lim_{t \rightarrow \infty} \frac{t^2}{3 t}\left(\text { form } \frac{\infty}{\infty}\right)=\ Lim_{t \rightarrow \infty} \frac{2 t}{3 e^{3 t}}\)
= \(\ Lim_{t \rightarrow \infty} \frac{2}{9 e^{3 t}}=\frac{2}{\infty}=0\)
Hence \(t^2\) is exponential order 3.
24. Prove that the Laplace transform of \(e^{t^2}\) does not exist.
Solution:
Let \(F(t)=e^{t^2} \text {. Then } \ Lim e^{-a t} F(t)=\ Lim_{t \rightarrow \infty} e^{-a t} e^{t^2}=\ Lim_{t \rightarrow \infty} e^{\left(t^{t^2}-a t\right)}\)
= \(\infty \text { i.e. } f(t)=e^{t^2}\) is not of exponential order. Hence its Laplace transform does not exist.
25. Find the Laplace transform of \(e^{2 t}+4 t^3-2 \sin 3 t+3 \cos 3 t\)
Solution:
L\(\left\{e^{2 t}+4 t^3-2 \sin 3 t+3 \cos 3 t\right\}\)
= \(\frac{1}{p-2}+4 \cdot \frac{3 !}{p^4}-2 \cdot \frac{3}{p^2+9}+3 \cdot \frac{p}{p^2+9}=\frac{1}{p-2}+\frac{24}{p^4}-\frac{6}{p^2+9}+\frac{3 p}{p^2+9} .\)
26. Find \(L\left(t^3+2 t^2-4 t+6\right)\).
Solution:
L\(\left(t^3+2 t^2-4 t+6\right)=L\left(t^3\right)+2 L\left(t^2\right)-4 L(t)+6 L(1)\)
= \(\frac{3 !}{p^4}+2 \cdot \frac{2 !}{p^3}-4 \cdot \frac{1}{p^2}+6 \cdot \frac{1}{p}=\frac{6}{p^4}+\frac{4}{p^3}-\frac{4}{p^2}+\frac{6}{p}\)
27. Find the Laplace transforms of \((t-2)^3\).
Solution:
L\(\left\{(t-2)^3\right\}=L\left\{t^3-6 t^2+12 t-8\right\}=L\left\{t^3\right\}-6 L\left\{t^2\right\}+12 L\{t\}-8 L\{1\}\)
= \(\frac{3 !}{p^4}-6 \frac{2 !}{p^3}+12 \frac{1 !}{p^2}-8 \frac{1}{p}=\frac{6}{p^4}-\frac{12}{p^3}+\frac{12}{p^2}-\frac{8}{p}\)
28. Find the Laplace transform of \(\left(t^2+1\right)^2\).
Solution:
L\(\left\{\left(t^2+1\right)^2\right\}=L\left\{t^4+2 t^2+1\right\}=L\left\{t^4\right\}+2 L\left\{t^2\right\}+L\{1\}\)
= \(\frac{4 !}{p^5}+2 \cdot \frac{2 !}{p^3}+\frac{1}{p}=\frac{24}{p^5}+\frac{4}{p^3}+\frac{1}{p}=\frac{1}{p^5}\left(p^4+4 p^2+24\right), p>0 .\)
29. Find \(L\left\{\left(\sqrt{t}+\frac{1}{\sqrt{t}}\right)^3\right\}\).
Solution:
L\(\left\{\left(\sqrt{t}+\frac{1}{\sqrt{t}}\right)^3\right\}=L\left\{t^{3 / 2}+3 t^{1 / 2}+3 t^{-1 / 2}+t^{-3 / 2}\right\}\)
= \(\frac{\Gamma(5 / 2)}{p^{5 / 2}}+3 \frac{\Gamma(3 / 2)}{p^{3 / 2}}+3 \frac{\Gamma(1 / 2)}{p^{1 / 2}}+\frac{\Gamma(-1 / 2)}{p^{-1 / 2}}\)
= \(\left(\frac{3}{2}\right)\left(\frac{1}{2}\right) \frac{\sqrt{\pi}}{p^{5 / 2}}+(3)\left(\frac{1}{2}\right) \frac{\sqrt{\pi}}{p^{3 / 2}}+(3) \frac{\sqrt{\pi}}{p^{1 / 2}}-\frac{2 \sqrt{\pi}}{p^{-1 / 2}}=\frac{\sqrt{\pi}}{4}\left[\frac{3}{p^{5 / 2}}+\frac{6}{p^{3 / 2}}+\frac{12}{p^{1 / 2}}-8 \sqrt{p}\right] .\)
30. Find the Laplace transform of \(\frac{e^{-a t}-1}{a}\)
Solution:
L\(\left\{\frac{e^{-a t}-1}{a}\right\}=\frac{1}{a} L\left\{e^{-a t}-1\right\}=\frac{1}{a}\left[L\left\{e^{-a t}\right\}-L\{1\}\right]=\frac{1}{a}\left[\frac{1}{p+a}-\frac{1}{p}\right]=-\frac{1}{p(p+a)}\)
31. Find the Laplace transform of \(e^{-t}-t^4+3 \sin (\sqrt{3} t)+4 \cos t\).
Solution:
L\(\left\{e^{-t}-t^4+3 \sin \sqrt{3} t+4 \cos t\right\}=\frac{1}{p+1}-\frac{4 !}{p^5}+3 \cdot \frac{\sqrt{3}}{p^2+3}+4 \cdot \frac{p}{p^2+1} .\)
32. Find \(L\left\{7 e^{2 t}+9 e^{-2 t}+5 \cos t+7 t^3+5 \sin 3 t+2\right\}\).
Solution:
L\(\left\{7 e^{2 t}+9 e^{-2 t}+5 \cos t+7 t^3+5 \sin 3 t+2\right\}\)
= \(7 L\left\{e^{2 t}\right\}+9 L\left\{e^{-2 t}\right\}+5 L\{\cos t\}+7 L\left\{t^3\right\}+5 L\{\sin 3 t\}+2 L\{1\}\)
= \(\frac{7}{p-2}+\frac{9}{p+2}+\frac{5 p}{p^2+1}+\frac{21}{p^4}+\frac{15}{p^2+9}+\frac{2}{p}\)
33. Find \(L\left\{e^{-t}+t^4+3 \sin 4 t-3 \sinh 2 t\right\}\).
Solution:
L\(\left\{e^{-t}+t^4+3 \sin 4 t-3 \sinh 2 t\right\}=L\left\{e^{-t}\right\}+L\left\{t^4\right\}+3 L\{\sin 4 t\}\)
–\(3 L\{\sinh 2 t\}\) \(\frac{1}{p+1}+\frac{4 !}{p^5}+3\left(\frac{4}{p^2+16}\right)-3\left(\frac{2}{p^2-4}\right)=\frac{1}{p+1}+\frac{24}{p^5}+\frac{12}{p^2+16}-\frac{6}{p^2-4}\)
34. Find \(L\left\{3 t^4-2 t^3+4 e^{-3 t}-2 \sin 5 t+3 \cos 2 t\right\}\).
Solution:
L\(\left\{3 t^4-2 t^3+4 e^{-3 t}-2 \sin 5 t+3 \cos 2 t\right\}\)
3\(L\left\{t^4\right\}-2 L\left\{t^3\right\}+4 L\left\{e^{-3 t}\right\}-2 L\{\sin 5 t\}+3 L\{\cos 2 t\}\)
= \(\frac{3(4 !)}{p^5}-\frac{2(3 !)}{p^4}+\frac{4}{p+3}-2 \frac{5}{p^2+5^2}+\frac{3 p}{p^2+2^2}\)
= \(\frac{72}{p^5}-\frac{12}{p^4}+\frac{4}{p+3}-\frac{10}{p^2+25}+\frac{3 p}{p^2+4}\)
35. Find \(L\left\{\left(5 e^{2 t}-3\right)^2\right\}\)
Solution:
L\(\left\{\left(5 e^{2 t}-3\right)^2\right\}=L\left\{25 e^{4 t}+9-30 e^{2 t}\right\}=25 L\left\{e^{4 t}\right\}-30 L\left\{e^{2 t}\right\}+9 L\{1\}\)
= \(\frac{25}{p-4}-\frac{30}{p-2}+\frac{9}{p}\).
36. Evaluate L{3 sin 2t – 5 cos 2t}.
Solution:
L\(\{3 \sin 2 t-5 \cos 2 t\}=3 L\{\sin 2 t\}-5 L\{\cos 2 t\}\)
= \(3\left(\frac{2}{p^2+2^2}\right)-5\left(\frac{p}{p^2+2^2}\right)=\frac{6-5 p}{p^2+4}\)
37. Find the Laplace transform of sin 2t cos t.
Solution:
L\(\{\sin 2 t \cos t\}=L\left\{\frac{1}{2}(\sin 3 t+\sin t)\right\}=\frac{1}{2}[L\{\sin 3 t\}+L\{\sin t\}]\)
= \(\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]=\frac{2\left(p^2+3\right)}{\left(p^2+1\right)\left(p^2+9\right)}\)
38. Find L(sin 2t cos 3t).
Solution:
L\((\sin 2 t \cos 3 t)=L\left(\frac{\sin 5 t-\sin t}{2}\right)=\frac{1}{2}[L(\sin 5 t)-L(\sin t)]\)
= \(\frac{1}{2}\left[\frac{5}{p^2+25}-\frac{1}{p^2+1}\right]\)
= \(\frac{1}{2}\left[\frac{5 p^2+5-p^2-25}{\left(p^2+25\right)\left(p^2+1\right)}\right]=\frac{1}{2}\left[\frac{4 p^2-20}{\left(p^2+25\right)\left(p^2+1\right)}\right]=\frac{2 p^2-10}{\left(p^2+25\right)\left(p^2+1\right)}\)
39. Find L[sin 3t cos 2t].
Solution:
⇒ \(\sin 3 t \cos 2 t=\frac{1}{2}(\sin 5 t+\sin t)\)
L\(\{\sin 3 t \cos 2 t\}=\frac{1}{2} L\{\sin 5 t+\sin t\}=\frac{1}{2}[L\{\sin 5 t\}+L\{\sin t\}]\)
= \(\frac{1}{2}\left[\frac{5}{p^2+25}-\frac{1}{p^2+1}\right]=\frac{1}{2}\left[\frac{5 p^2+5-p^2-25}{\left(p^2+25\right)\left(p^2+1\right)}\right]=\frac{1}{2}\left[\frac{4 p^2-20}{\left(p^2+25\right)\left(p^2+1\right)}\right]\)
= \(\frac{2\left(p^2-5\right)}{\left(p^2+25\right)\left(p^2+1\right)}\)
40. Find the Laplace transform of cos 3t sin 5t.
Solution:
L\(\{\sin 5 t \cos 3 t\}=L\left\{\frac{1}{2}(\sin 8 t+\sin 2 t)\right\}=\frac{1}{2}[L\{\sin 8 t\}+L\{\sin 2 t\}]\)
= \(\frac{1}{2}[\dot{L}\{\sin 5 t\}+L\{\sin t\}]=\frac{1}{2}\left[\frac{8}{p^2+64}+\frac{2}{p^2+4}\right]\)
= \(\frac{1}{2}\left[\frac{8 p^2+32+2 p^2+128}{\left(p^2+64\right)\left(p^2+4\right)}\right]\)
= \(\frac{1}{2}\left[\frac{10 p^2+160}{\left(p^2+64\right)\left(p^2+4\right)}\right]=\frac{5 p^2+80}{\left(p^2+64\right)\left(p^2+4\right)}\)
41. Find the Laplace transform of cos t cos 2t cos 3t.
Solution:
L\(\{\cos t \cos 2 t \cos 3 t\}=L\left\{\frac{\cos t}{2}(\cos 5 t+\cos t)\right\}\)
= \(L\left\{\frac{1}{4}\left[2 \cos 5 t \cos t+2 \cos ^2 t\right]\right\}\)
= \(\frac{1}{4} L\{\cos 6 t+\cos 4 t+\cos 2 t+1\}=\frac{1}{4}[L\{\cos 6 t\}\) \(+L\{\cos 4 t\}+L\{\cos 2 t\}+L\{1\}]\)
= \(\frac{1}{4}\left[\frac{p}{p^2+6^2}+\frac{p}{p^2+4^2}+\frac{p}{p^2+2^2}+\frac{1}{p}\right]=\frac{1}{4}\left[\frac{1}{p}+\frac{p}{p^2+4}+\frac{p}{p^2+16}+\frac{p}{p^2+36}\right]\)
42. Find the Laplace transform of \((\sin t+\cos t)^2\).
Solution:
⇒ \((\sin t+\cos t)^2=\sin ^2 t+\cos ^2 t+2 \sin t \cos t=1+\sin 2 t\)
∴ \(L\left\{(\sin t+\cos t)^2\right\}=L\{1+\sin 2 t\}=\frac{1}{p}+\frac{2}{p^2+4}=\frac{p^2+2 p+4}{p\left(p^2+4\right)}\)
43. Find \(L\left[(\sin t-\cos t)^2\right]\).
Solution:
⇒ \((\sin t-\cos t)^2=\sin ^2 t+\cos ^2 t-2 \sin t \cos t=1-\sin 2 t\)
∴ \(L\left\{(\sin t-\cos t)^2\right\}=L\{1-\sin 2 t\}=\frac{1}{p}-\frac{2}{p^2+4}=\frac{p^2-2 p+4}{p\left(p^2+4\right)}\)
44. Find \(L\left(\sin ^2 4 t\right)\).
Solution:
L\(\left(\sin ^2 4 t\right)=L\left[\frac{1-\cos 8 t}{2}\right]=\frac{1}{2} L(1)-\frac{1}{2} L(\cos 8 t)\)
= \(\frac{1}{2} \cdot \frac{1}{p}-\frac{1}{2} \cdot \frac{p}{p^2+64}=\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+64}\right]=\frac{1}{2}\left[\frac{p^2+64-p^2}{p\left(p^2+64\right)}\right]=\frac{32}{p\left(p^2+64\right)}\)
45. Find \(L\left\{\sin ^2 a t\right\}\).
Solution:
L\(\left\{\sin ^2 a t\right\}=L\left\{\frac{1-\cos 2 a t}{2}\right\} \doteq \frac{1}{2} L\{1\}-\frac{1}{2} L\{\cos 2 a t\}\)
= \(\frac{1}{2} \frac{1}{p}-\frac{1}{2} \frac{p}{p^2+4 a^2}=\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+4 a^2}\right]=\frac{1}{2}\left[\frac{4 a^2}{p\left(p^2+4 a^2\right)}\right]=\frac{2 a^2}{p\left(p^2+4 a^2\right)}\)
46. Find \(L\left\{\sin ^2 a t+\left(t^2+1\right)^2\right\}\).
Solution:
L\(\left\{\sin ^2 a t+\left(t^2+1\right)^2\right\}=L\left\{\frac{1-\cos 2 a t}{2}+t^4+2 t^2+1\right\}=L\left\{\frac{3}{2}-\frac{1}{2} \cos 2 a t+t^4+2 t^2\right\}\)
= \(\frac{3}{2} L\{1\}-\frac{1}{2} L\{\cos 2 a t\}+L\left\{t^4\right\}+2 L\left\{t^2\right\}\)
= \(\frac{3}{2}\left(\frac{1}{p}\right)-\frac{1}{2}\left(\frac{p}{p^2+4 a^2}\right)+\frac{4 !}{p^5}+2\left(\frac{2 !}{p^3}\right)=\frac{3}{2 p}-\frac{p}{2\left(p^2+4 a^2\right)}+\frac{24}{p^5}+\frac{4}{p^3}\)
47. Find the Laplace transform of \(\cos ^3 2 t\).
Solution:
cos 6 t = \(\cos 3(2 t)=4 \cos ^3 2 t-3 \cos 2 t\)
∴ \(\cos ^3 2 t=\frac{1}{4}(3 \cos 2 t+\cos 6 t)\)
Hence \(L\left\{\cos ^3 2 t\right\}=L\left\{\frac{1}{4}(3 \cos 2 t+\cos 6 t)\right\}=\frac{3}{4} L\{\cos 2 t\}+\frac{1}{4} L\{\cos 6 t\}\)
= \(\frac{3}{4} \cdot \frac{p}{p^2+4}+\frac{1}{4} \cdot \frac{p}{p^2+36}=\frac{p}{4}\left(\frac{3}{p^2+4}+\frac{1}{p^2+36}\right)=\frac{p}{4}\left[\frac{4 p^2+112}{\left(p^2+4\right)\left(p^2+36\right)}\right]\)
= \(\frac{p\left(p^2+28\right)}{\left(p^2+4\right)\left(p^2+36\right)}\)
48. Find Laplace Transform of \(\sin ^3 2 t\).
Solution:
L\(\left\{\sin ^3 2 t\right\}=L\left\{\frac{3 \sin 2 t-\sin 6 t}{4}\right\}=\frac{3}{4} L\{\sin 2 t\}-\frac{1}{4} L\{\sin 6 t\}=\frac{3}{4} \frac{2}{p^2+4}-\frac{1}{4} \frac{6}{p^2+36}\)
= \(\frac{3}{2}\left[\frac{1}{p^2+4}-\frac{1}{p^2+36}\right]=\frac{3}{2}\left[\frac{p^2+36-p^2-4}{\left(p^2+4\right)\left(p^2+36\right)}\right]=\frac{48}{\left(p^2+4\right)\left(p^2+36\right)}\)
49. Find \(L\left(\cos ^3 3 t\right)\).
Solution:
We know that \(cos 9 t=\cos 3(3 t)=4 \cos ^3 3 t-3 \cos 3 t\)
Hence \(\cos ^3 3 t=\frac{1}{4} \cos 9 t+\frac{3}{4} \cos 3 t\)
L\(\left(\cos ^3 3 t\right)=\frac{1}{4} L(\cos 9 t)+\frac{3}{4} L(\cos 3 t)=\frac{1}{4} \frac{p}{p^2+81}+\frac{3}{4} \cdot \frac{p}{p^2+9}\)
= \(\frac{p\left(p^2+9+3 p^2+243\right)}{4\left(p^2+81\right)\left(p^2+9\right)}=\frac{p\left(4 p^2+252\right)}{4\left(p^2+81\right)\left(p^2+9\right)}=\frac{p\left(p^2+63\right)}{\left(p^2+81\right)\left(p^2+9\right)}\)
50. Find the Laplace transform of \((\sin t-\cos t)^3\).
Solution:
(\(sin t-\cos t)^3=\sin ^3 t-3 \sin ^2 t \cos t+3 \sin t \cos ^2 t-\cos ^3 t\)
= \(\sin ^3 t-3 \cos t\left(1-\cos ^2 t\right)+3 \sin t\left(1-\sin ^2 t\right)-\cos ^3 t\)
= \(\sin ^3 t-3 \cos t+3 \cos ^3 t+3 \sin t-3 \sin ^3 t-\cos ^3 \)
= \(2 \cos ^3 t-3 \cos t+3 \sin t-2 \sin ^3 t\)
= \(\frac{3 \cos t+\cos 3 t}{2}-3 \cos t+3 \sin t-\frac{3 \sin t-\sin 3 t}{2}\)
= \(\frac{1}{2} \cos 3 t-\frac{3}{2} \cos t+\frac{3}{2} \sin t+\frac{1}{2} \sin 3 t\)
L\(\left\{(\sin t-\cos t)^3\right\}=\frac{1}{2} L\{3 \sin t-3 \cos t+\sin 3 t+\cos 3 t\}\)
= \(\frac{1}{2}\left[\frac{3}{p^2+1}-\frac{3 p}{p^2+1}+\frac{3}{p^2+9}+\frac{p}{p^2+9}\right]\)
51. Find the Laplace transform of \(\cosh ^2 2 t\).
Solution:
L\(\left\{\cosh ^2 2 t\right\}=L\left\{\frac{1}{2}(1+\cosh 4 t)\right\}=\frac{1}{2}[L\{1\}+L\{\cosh 4 t\}]\)
= \(\frac{1}{2}\left[\frac{1}{p}+\frac{p}{p^2-16}\right]=\frac{p^2-8}{p\left(p^2-16\right)}\)
52. Find the Laplace transform of \(L\left\{\sinh ^2 3 t\right\}\).
Solution:
⇒ \(\sinh ^2(3 t)=\left(\frac{e^{3 t}-e^{-3 t}}{2}\right)^2=\frac{e^{9 t}+e^{-9 t}-2}{4}\)
Hence \(L\left\{\sinh ^2 3 t\right\}=\frac{1}{4}\left[L\left\{e^{9 t}\right\}+L\left\{e^{-9 t}\right\}-2 L\{1\}\right]=\frac{1}{4}\left[\frac{1}{p-9}+\frac{1}{p+9}-\frac{2}{p}\right] \text {. }\)
53. Find the Laplace transform of \(\sin h^3 2 t\).
Solution:
L\(\left\{\sin h^3 2 t\right\}=L\left\{\left(\frac{e^{2 t}-e^{-2 t}}{2}\right)^3\right\}=L\left\{\frac{1}{8}\left(e^{6 t}-3 e^{2 t}+3 e^{-2 t}-e^{-6}\right)\right\}\)
= \(=\frac{1}{8} \cdot \frac{1}{p-6}-\frac{3}{8} \cdot \frac{1}{p-2}+\frac{3}{8} \cdot \frac{1}{p+2}-\frac{1}{8} \cdot \frac{1}{p+6}\)
= \(\frac{1}{8}\left(\frac{1}{p-6}-\frac{1}{p+6}\right)-\frac{3}{8}\left(\frac{1}{p-2}-\frac{1}{p+2}\right)\)
= \(\frac{1}{8}\left(\frac{12}{p^2-36}-\frac{12}{p^2-4}\right)=\frac{48}{\left(p^2-4\right)\left(p^2-36\right)}\)
54. Find the Laplace transform of \(\cosh ^3 2 t\).
Solution:
⇒ \(\cosh 6 t=4 \cosh ^3 2 t-3 \cosh 2 t \Rightarrow \cosh ^3 2 t=\frac{\cosh 6 t+3 \cosh 2 t}{4}\)
∴ \(L\left(\cosh ^3 2 t\right)=\frac{1}{4}\left[L\{\cosh 6 t\}+3 L\{\cosh 2 t\}=\frac{1}{4}\left[\frac{p}{p^2-36}+\frac{3 p}{p^2-4}\right]\right.\)
= \(\frac{p}{4}\left[\frac{1}{p^2-36}+\frac{3}{p^2-4}\right]=\frac{p\left(p^2-28\right)}{\left(p^2-4\right)\left(p^2-36\right)}\)
55. Prove that \(L\left\{e^{a t} \sinh b t\right\}=\frac{b}{(p-a)^2-b^2}\).
Solution:
L\(\left\{e^{a t} \sinh b t\right\}=\int_0^{\infty} e^{-p t} e^{a t} \sinh b t d t=\int_0^{\infty} e^{-(p-a) t} \sinh b t d t\)
= \(\int_0^{\infty} e^{-(p-a) t}\left(\frac{e^{b t}-e^{-b t}}{2}\right) d t=\frac{1}{2} \int_0^{\infty}\left(e^{-(p-a-b) t}-e^{-(p-a+b) t}\right) d t\)
= \(\frac{1}{2}\left[\frac{e^{-(p-a-b) t}}{-(p-a-b)}-\frac{e^{-(p-a+b) t}}{-(p-a+b)}\right]_b^p=\frac{1}{2}\left[\frac{1}{p-a-b}-\frac{1}{p-a+b}\right]\)
= \(\frac{1}{2}\left[\frac{2 b}{(p-a)^2-b^2}\right]=\frac{b}{(p-a)^2-b^2}\)
56. Prove that \(L\left\{e^{a t} \cosh b t\right\}=\frac{p-a}{(p-a)^2-b^2}\).
Solution:
L\(\left\{e^{a t} \cosh b t\right\}=\int_0 e^{-p t} e^{a t} \cosh b t d t=\int_0^{\infty} e^{-(p-a) t} \frac{e^{b t}+e^{-b t}}{2} d t\)
= \(\frac{1}{2} \int_0^{\infty}\left[e^{-(p-a-b) t}+e^{-(p-a+b) t}\right] d t=\frac{1}{2}\left[\frac{e^{-(p-a-b) t}}{-(p-a-b)}+\frac{e^{-(p-a+b) t}}{-(p-a+b)}\right]_0^{\infty}\)
= \(\frac{1}{2}\left[\frac{1}{p-a-b}+\frac{1}{p-a+b}\right]=\frac{p-a}{(p-a)^2-b^2}\)
57. Find the Laplace transform of f(t)=|t-1|+|t+1|,t≥0.
Solution:
0<t<1\(\Rightarrow|t-1|=1-t \text { and }|t+1|=t+1\)
∴ \(F(t)=(1-t)+(t+1)=2, \text { when } 0<t<1 \rightarrow(1)\)
When \(t>1 \Rightarrow|t-1|=t-1 \text {. }\)
∴ \(F(t)=(t-1)+(t+1)=2 t \text {, when } t>1\)
Thus \(L\{F(t)\}=\int_0^{\infty} F(t) e^{-p t} d t=\int_0^1 2 e^{-p t} d t+\int_1^{\infty} 2 t e^{-p t} d t=\left(\frac{e^{-p t}}{-p}\right)_0^1+2\left[t\left(\frac{e^{-p t}}{-p}\right)-\left(\frac{e^{-p t}}{p^2}\right)\right]_1^{\infty}\)
= \(2\left(\frac{e^{-p}}{-p}+\frac{1}{p}\right)+2\left(\frac{e^{-p}}{p}+\frac{e^{-p}}{p^2}\right)=\frac{2}{p}\left(1+\frac{e^{-p}}{p}\right) .\)
58. Find the Laplace transform of \(f(t)=\left\{\begin{array}{c}
e^t \text { when } 0<t<1 \\
0 \text { when } t>1
\end{array}\right.\).
Solution:
L[f(t)]=\(\int_0^{\infty} e^{-p t} f(t) d t=\int_0^1 e^{-p t} f(t) d t+\int_1^{\infty} e^{-p t} f(t) d t\) \(\int_0^1 e^{-p t} e^t d t+\int_1^{\infty} e^{-p t}(0) d t\)
= \(\int_0^1 e^{(1-p) t} d t=\left[\frac{e^{(1-p) t}}{1-p}\right]_0^1=\frac{1}{1-p}\left(e^{1-p}-1\right)\)
59. Obtain the Laplace transform of the function \(F(t)=\left\{\begin{array}{cl}
(t-1)^2, & t>1 \\
0, & 0<t<1
\end{array}\right.\).
Solution:
By defination, \(L\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^1 e^{-p t} F(t) d t+\int_1^{\infty} e^{-p t} F(t) d t\)
= \(\int_0^1 e^{-p t} \cdot 0 d t+\int_1^{\infty} e^{-p t}(t-1)^2 d t=\int_1^{\infty} e^{-p t}(t-1)^2 d t\)
= \(\left[(t-1)^2\left(\frac{e^{-p t}}{-p}\right)-2(t-1)\left(\frac{e^{-p t}}{p^2}\right)-2\left(\frac{e^{-p t}}{p^3}\right)\right]_1^{\infty}\)
= \(\left[(0-0-0)-\left\{0-0-\frac{2}{p^3} e^{-p}\right\}\right]=\frac{2}{p^3} e^{-p}\)
60. Find the Laplace transform of \(F(t)=\left\{\begin{array}{cl}
\sin t, & 0<t<\pi \\
0 & t>\pi
\end{array}\right.\)
Solution:
L\(\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^\pi e^{-p t} F(t) d t+\int_\pi^{\infty} e^{-p t} F(t) d t\)
= \(\int_0^\pi e^{-p t} \sin t d t+\int_\pi^{\infty} e^{-p t}(0) d t\)
= \(\int_0^\pi e^{-p t} \sin t d t=\left[\frac{e^{-p t}}{p^2+1}(-p \sin t-\cos t)\right]_0^\pi=\frac{1+e^{-\pi p}}{p^2+1}\)
61. Find the Laplace transform of F(t) defined as \(F(t)= \begin{cases}e^t & \text { when } 0<t<5 \\ 3 & \text { when } t>5\end{cases}\)
Solution:
L\(\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^5 e^{-p t} F(t) d t+\int_5^{\infty} e^{-p t} F(t) d t\)
= \(\int_0^5 e^{-p t} e^t d t+\int_5^{\infty} e^{-p t} 3 d t=\int_0^5 e^{-(p-1) t} d t+3 \int_5^{\infty} e^{-p t} d t=\left[\frac{e^{-(p-1) t}}{-(p-1)}\right]_0^5+3\left(\frac{e^{-p}}{-p}\right)_5^{\infty}\)
= \(\left[\frac{-e^{-5(p-1)}}{p-1}+\frac{1}{p-1}\right]+3 \frac{e^{-5 p}}{p}=\frac{1-e^{-5(p-1)}}{p-1}+\frac{3}{p} e^{-5 p}\)
62. Find the Laplace transform of \(F(t)= \begin{cases}2 t, & 0<t<5 \\ 1, & t>5\end{cases}\).
Solution:
L\([f(t)]=\int_0^{\infty} e^{-p t} f(t) d t=\int_0^5 e^{-p t} f(t) d t+\int_5^{\infty} e^{-p t} f(t) d t=\int_0^5 e^{-p t} 2 t d t+\int_5^{\infty} e^{-p t}(1) d t\)
= \(\left(\frac{2 t e^{-p t}}{-p}\right)_0^5-\int_0^5 2 \frac{e^{-p t}}{-p} d t+\left(\frac{e^{-p t}}{-p}\right)_5^{\infty}=\frac{-10}{p} e^{-5 p}-\left(\frac{2 e^{-p t}}{p^2}\right)_0^5+\frac{e^{-5 p}}{p}\)
= \(-\frac{9}{p} e^{-5 p}-\frac{2}{p^2} e^{-5 p}+\frac{2}{p^2}=\frac{2}{p^2}\left(1-e^{-5 p}\right)-\frac{9}{p} e^{-5 p}\)
63. Find L{F(t)} where \(F(t)= \begin{cases}0, & 0<t<2 \\ 4, & t>2\end{cases}\)
Solution:
L\(\{\hat{F}(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_2^{\infty} 4 e^{-p t} d t=\left[\frac{4 e^{-p t}}{-p}\right]_2^{\infty}=\frac{4}{p} e^{-2 p}\)
64. Find the Laplace Transform of F(t) defined as \(F(t)= \begin{cases}1, & 0<t<2 \\ 2, & 2<t<4 \\ 3, & 4<t<6 \\ 0, & t>6\end{cases}\)
Solution:
L\(\{f(t)\}=\int_0^2 e^{-p t}(1) d t+\int_2^4 e^{-p t} 2 d t+\int_4^6 e^{-p t} 3 d t+\int_6^{\infty} e^{-p t}(0) d t\)
=\(\left(\frac{e^{-p t}}{-p}\right)_0^2+2\left(\frac{e^{-p t}}{-p}\right)_2^4+3\left(\frac{e^{-p t}}{-p}\right)_4^6=-\frac{1}{p}\left(e^{-2 p}-1\right)-\frac{2}{p}\left(e^{-4 p}-e^{-2 p}\right)-\frac{3}{p}\left(e^{-6 p}-e^{-4 p}\right)\)
= \(\frac{1}{p}\left(1+e^{-2 p}+e^{-4 p}-3 e^{-6 p}\right)\)
65. Find L{F(t)} if \(F(t)= \begin{cases}0, & 0<t<1 \\ t, & 1<t<2 \\ 0, & t>2\end{cases}\)
Solution:
L\(\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^1 e^{-p t} F(t) d t+\int_1^2 e^{-p t} \cdot F(t) d t+\int_2^{\infty} e^{-p t} F(t) d t\)
= \(\int_0^1 e^{-p t}(0) d t+\int_1^2 e^{-p t}(t) d t+\int_2^{\infty} e^{-p t}(0) F(t) d t=\int_1^2 t e^{-p t} d t\)
= \(\left[t\left(\frac{e^{-p t}}{-p}\right)-1\left(\frac{e^{-p t}}{p^2}\right)\right]_1^2=-\left[\frac{t}{p} e^{-p t}+\frac{1}{p^2} e^{-p t}\right]_1^2=-\left[\frac{2}{p} e^{-2 p}+\frac{1}{p^2} e^{-2 p}\right]+\left[\frac{1}{p} e^{-p}+\frac{1}{p^2} e^{-p}\right]\)
= \(-e^{-2 p}\left(\frac{2}{p}+\frac{1}{p^2}\right)+e^{-p}\left(\frac{1}{p}+\frac{1}{p^2}\right)\)
66. Find L{f(t)} where \(F(t)=\left\{\begin{array}{cc}
e^{t-a}, & t>a \\
0, & t<a
\end{array}\right.\)
Solution:
L\(\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^a e^{-p t} F(t) d t+\int_a^{\infty} e^{-p t} F(t) d t=\int_0^a e^{-p t}(0) d t+\int_a^{\infty} e^{-p t} e^{t-a} d t\)
= \(\int_a^{\infty} e^{-a} e^{-(p-1) t} d t=e^{-a}\left[\frac{e^{-(p-1) t}}{-(p-1)}\right]_a^{\infty}=e^{-a}\left[\frac{-e^{-(p-1) a}}{-(p-1)}\right]=\frac{e^{-p a}}{p-1},(p>1)\)
67. Find the Laplace transform of the function \(F(t)= \begin{cases}4, & 0<t<1 \\ 3, & t>1\end{cases}\)
Solution:
L\(\{F(t)\}=\int_0^{\infty} e^{-p t} F(t) d t=\int_0^1 F(t) e^{-p t} d t+\int_1^{\infty} F(t) e^{-p t} d t=\int_0^1 4 e^{-p t} d t+\int_1^{\infty} 3 e^{-p t} d t\)
= \(4\left[\frac{e^{-p t}}{-p}\right]_0^1+3\left[\frac{e^{-p t}}{-p}\right]_1^{\infty}=4\left[\frac{e^{-p}-e^0}{-p}\right]+3\left[\frac{e^{-\infty}-e^{-p}}{-p}\right]=4\left[\frac{e^{-p}-1}{-p}\right]+3\left[\frac{-e^{-p}}{-p}\right]\)
= \(\frac{4 e^{-p}-4-3 e^{-p}}{-p}=\frac{-e^{-p}+4}{p}=\frac{4-e^{-p}}{p}\)
68. Show that \(L\left\{\frac{1}{\sqrt{\pi t}}\right\}=\frac{1}{\sqrt{p}}\).
Solution:
L\(\left\{\frac{1}{\sqrt{\pi} t}\right\}=\frac{1}{\sqrt{\pi}} L\left\{\frac{1}{\sqrt{t}}\right\}=\frac{1}{\sqrt{\pi}} L\left\{t^{-1 / 2}\right\}=\frac{1}{\sqrt{\pi}} \frac{\Gamma(-1 / 2+1)}{p^{-1 / 2+1}}=\frac{1}{\sqrt{\pi}} \frac{\Gamma(1 / 2)}{p^{1 / 2}}=\frac{1}{\sqrt{\pi}} \cdot \frac{\sqrt{\pi}}{\sqrt{p}}=\frac{1}{\sqrt{p}}\)
69. Using expansion \(\sin x=x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\frac{x^7}{7 !}+\ldots\), show that \(L\{\sin \sqrt{t}\}=\frac{\sqrt{\pi}}{2 p^{3 / 2}} e^{-1 / 4 p}\).
Solution:
sin \(\sqrt{t}=\sqrt{t}-\frac{(\sqrt{t})^3}{3!}+\frac{(\sqrt{t})^5}{5!}-\frac{(\sqrt{t})^7}{7!}+\cdots=t^{1 / 2}-\frac{t^{3 / 2}}{3!}+\frac{t^{5 / 2}}{5!}-\frac{t^{7 / 2}}{7!}+\cdots\)
∴ \(L\{\sin \sqrt{t}\}=L\left\{t^{1 / 2}\right\}-\frac{1}{3!} L\left\{t^{3 / 2}\right\}+\frac{1}{5!} L\left\{t^{5 / 2}\right\}-\frac{1}{7!} L\left\{t^{7 / 2}\right\}+\cdots\)
= \(\frac{\Gamma\left(\frac{3}{2}\right)}{p^{3 / 2}}-\frac{1}{3!} \frac{\Gamma\left(\frac{5}{2}\right)}{p^{5 / 2}}+\frac{1}{5!} \frac{\Gamma\left(\frac{7}{2}\right)}{p^{7 / 2}}-\frac{1}{7!} \frac{\Gamma\left(\frac{9}{2}\right)}{p^{9 / 2}}+\cdots\)
therefore \(L\left\{t^n\right\}=\frac{\Gamma(n+1)}{p^{n+1}}\)
= \(\frac{\frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{3 / 2}}-\frac{1}{3!} \frac{\frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{5 / 2}}+\frac{1}{5!} \frac{\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{7 / 2}}-\frac{1}{7!} \frac{\frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \Gamma\left(\frac{1}{2}\right)}{p^{9 / 2}}+\cdots\)
= \(\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left[1-\frac{3}{2 \cdot 3!p}+\frac{5 \cdot 3}{2 \cdot 2 \cdot 5!p^2}-\frac{7 \cdot 5 \cdot 3}{2 \cdot 2 \cdot 2 \cdot 7!p^3}+\cdots\right]\) (because \(\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\)
= \(\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left[1-\frac{1}{2^2} p+\frac{1}{\left(2^2 p\right)^2 2!}-\frac{1}{\left(2^2 p\right)^3 3!}+\cdots\right]=\frac{\sqrt{\pi}}{2 p^{3 / 2}}\left(e^{-1 /\left(2^2 p\right)}\right)=\frac{\sqrt{\pi}}{2 p^{3 / 2}} e^{-1 / 4 p}\)
70. State and prove first translation or shifting theorem
Solution:
Statement: If L[F(t)]=f(p), then
1) \(L\left[e^{-a t} F(t)\right]=f(p+a)\)
2) \(L\left[e^{a t} F(t)\right]=f(p-a)\)
Proof: By definition, \(L[F(t)]=\int_0^{\infty} e^{-p t} \cdot F(t) d t\)
1) \(L\left[e^{-a t} \cdot \dot{F}(t)\right]=\int_0^{\infty} e^{-p t} e^{-a t} F(t) d t=\int_0^{\infty} e^{-(p+a) t} F(t) d t=\int_0^{\infty} e^{-u t} F(t) d t\)
where u=p+a. Hence \(L\left[e^{-a t} F(t)\right]=f(p+a)\)
2) \(L\left[e^{a t} F(t)\right]=\int_0^{\infty} e^{- p t} e^{a t} F(t) d t=\int_0^{\infty} e^{-(p-a) t} F(t) d t=\int_0^{\infty} e^{-u t} F(t) d t\)
where u=p-a =f(u)=f(p-a). Hence \(L\left[e^{a t} F(t)\right]=f(p-a)\).
71. Show that
1) \(L\left[e^{-a t} \cos b t\right]=\frac{p+a}{(p+a)^2+b^2}\)
2) \(L\left[e^{a t} \cos b t\right]=\frac{p-a}{(p-a)^2+b^2}\)
Solution:
1) By first shifting theorem, \(L[\cos b t]=\left[\frac{p}{p^2+b^2}\right] \Rightarrow L\left[e^{-a t} \cos b t\right]=\left[\frac{p}{p^2+b^2}\right]_{p \rightarrow p+a}=\frac{p+a}{(p+a)^2+b^2}\).
2) By first shifting theorem, \(L[\cos b t]=\left[\frac{p}{p^2+b^2}\right] \Rightarrow L\left[e^{a t} \cos b t\right]=\left[\frac{p}{p^2+b^2}\right]_{p \rightarrow p-a}=\frac{p-a}{(p-a)^2+b^2}\).
72. Show that
1) \(L\left[e^{-a t} \sin b t\right]=\frac{b}{(p+a)^2+b^2}\)
2) \(L\left[e^{a t} \sin b t\right]=\frac{b}{(p-a)^2+b^2}\)
Solution:
1) By first shifting theorem, \(L[\sin b t]=\left[\frac{b}{p^2+b^2}\right] \Rightarrow L\left[e^{-a t} \sin b t\right]=\left[\frac{b}{p^2+b^2}\right]_{p \rightarrow p+a}=\frac{b}{(p+a)^2+b^2} .\).
2) By first shifting theorem, \(L[\sin b t]=\left[\frac{b}{p^2+b^2}\right] \Rightarrow L\left[e^{a t} \sin b t\right]=\left[\frac{b}{p^2+b^2}\right]_{p \rightarrow p-a}=\frac{b}{(p-a)^2+b^2}\).
73. Show that
1) \(L\left[e^{-a t} t^n\right]=\frac{n !}{(p+a)^{n+1}}\)
2) \(L\left[e^{a t} t^n\right]=\frac{n !}{(p-a)^{n+1}}\)
Solution:
1) By first shifting theorem, \(L\left[t^n\right]=\frac{n !}{p^{n+1}} \Rightarrow L\left[e^{-a t} t^n\right]=\left[\frac{n !}{p^{n+1}}\right]_{p \rightarrow p+a}=\frac{n !}{(p+a)^{n+1}}\).
2) By first shifting theorem, \(L\left[t^n\right]=\frac{n !}{p^{n+1}} \Rightarrow L\left[e^{a t} t^n\right]=\left[\frac{n !}{p^{n+1}}\right]_{p \rightarrow p-a}=\frac{n !}{(p-a)^{n+1}}\).
74. Show that
1) \(L\left[e^{-a t} \cosh b t\right]=\frac{p+a}{(p+a)^2-b^2}\)
2) \(L\left[e^{a t} \cosh b t\right]=\frac{p-a}{(p-a)^2-b^2}\)
Solution:
1) By first shifting theorem, L\(L[\cosh b t]=\frac{p}{p^2-b^2} \Rightarrow L\left[e^{-a t} \cosh b t\right]=\left[\frac{p}{p^2-b^2}\right]_{p \rightarrow p+a}=\frac{p+a}{(p+a)^2-b^2}\).
2) By first shifting theorem, \(L[\cosh b t]=\frac{p}{p^2-b^2} \Rightarrow L\left[e^{a t} \cos b t\right]=\left[\frac{p}{p^2-b^2}\right]_{p \rightarrow p-a}=\frac{p-a}{(p-a)^2-b^2}\).
75. Show that
1) \(L\left[e^{-a t} \sinh b t\right]=\frac{b}{(p+a)^2-b^2}\)
2) \(L\left[e^{a t} \sinh b t\right]=\frac{b}{(p-a)^2-b^2}\)
Solution:
1) By first shifting theorem, \(L\left[[\sinh b t]=\frac{b}{p^2-b^2} \Rightarrow L\left[e^{-a t} \sinh \ b t\right]=\left[\frac{b}{p^2-b^2}\right]_{p \rightarrow p+a}=\frac{b}{(p+a)^2-b^2}\right.\).
2) By first shifting theorem, \(L[\sinh b t]=\frac{b}{p^2-b^2} \Rightarrow L\left[e^{a t} \sinh b t\right]=\left[\frac{b}{p^2-b^2}\right]_{p \rightarrow p-a}=\frac{b}{(p-a)^2+b^2}\).
76. Find the Laplace transform of \(e^{-t} \cos 2 t\).
Solution:
L\(\{\cos 2 t\}=\frac{p}{p^2+4}\left[\ L(\cos a t)=\frac{p}{p^2+a^2}\right]\)
Now Applying the First shifting theorem, we get
L\(\left\{e^{-t} \cos 2 t\right\}=[L\{\cos 2 t\}]_{p \rightarrow p+1}=\left[\frac{p}{p^2+4}\right]_{p \rightarrow p+1}=\frac{p+1}{(p+1)^2+4}=\frac{p+1}{p^2+2 p+5}\)
77. Find the Laplace transform of \(e^{-3 t} \sin t\).
Solution:
L\(\{\sin t\}=\frac{1}{p^2+1} \text {. So } L\left\{e^{-3 t} \sin t\right\}=\left(\frac{1}{p^2+1}\right)_{p \rightarrow p+3}=\frac{1}{(p+3)^2+1}=\frac{1}{p^2+6 p+10}\)
78. Find \(L\left[e^{-2 t} \sin 3 t\right]\).
Solution:
We know that, \(L(\sin 3 t)=\frac{3}{p^2+9}\)
L\(\left(e^{-2 t} \sin 3 t\right)=\left(\frac{3}{p^2+9}\right)_{p \rightarrow p+2}=\frac{3}{(p+2)^2+9}=\frac{3}{p^2+4 p+13}\)
79. Find \(L\left\{e^{3 t} \sin 4 t\right\}\).
Solution:
We have, \(L\{\sin 4 t\}=\frac{4}{p^2+16}=f(p)\)
Using First Shifting Theorem, \(L\left\{e^{3 t} \sin 4 t\right\}=6 f(p-3)=\frac{4}{(p-3)^2+16}=\frac{4}{p^2-6 p+25}\)
80. Find \(L\left\{e^{4 t} \cos 5 t\right\}\).
Solution:
We know \(L[\cos 5 t]=\left[\frac{p}{p^2+5^2}\right] \text {. By first shifting theorem, }\)
L\(\left[e^{4 t} \cos 5 t\right]=\left[\frac{p}{p^2+5^2}\right]_{p \rightarrow p-4}=\frac{p-4}{(p-4)^2+5^2}=\frac{p-4}{p^2-8 p+41} .\)
81. Find the Laplace transform of \(e^{-3 t}(2 \cos 5 t-3 \sin 5 t)\).
Solution:
We have \(L(2 \cos 5 t-3 \sin 5 t)=2 \cdot \frac{p}{p^2+35}-3 \cdot \frac{5}{p^2+25}=\frac{2 p-15}{p^2+25}\)
Now Applying the First Shifting Theorem, we have
L\(\left\{e^{-3 t}(2 \cos 5 t-3 \sin 5 t)\right\}=\left(\frac{2 p-15}{p^2+25}\right)_{p \rightarrow p+3}=\frac{2(p+3)-15}{(p+3)^2+25}=\frac{2 p-9}{p^2+6 p+34} .\).
82. Find \(L\left\{e^{-2 t}(3 \cos 6 t-5 \sin 6 t)\right\}\).
Solution:
By First Shifting Theorem,
L\(\left\{e^{-2 t}(3 \cos 6 t-5 \sin 6 t)\right\}=[L\{3 \cos 6 t-5 \sin 6 t\}]_{p \rightarrow p+2}\)
= \(\left[\frac{3 p}{p^2+36}-\frac{30}{p^2+36}\right]_{p \rightarrow p+2}=\frac{3(p+2)-30}{(p+2)^2+36}=\frac{3 p-24}{p^2+4 p+40}\)
83. Find \(L\left\{e^{-t} \cos ^2 t\right\}\).
Solution:
We have \(L\left\{\cos ^2 t\right\}=L\left\{\frac{1+\cos 2 t}{2}\right\}=\frac{1}{2}[L\{1\}+L\{\cos 2 t\}]=\frac{2\left(p^2+2\right)}{p\left(p^2+4\right)}=f(p)\)
By First Shifting Theorem,
L\(\left\{e^{-t} \cos ^2 t\right\}=f(p+1)=\left[\frac{2\left(p^2+2\right)}{p\left(p^2+4\right)}\right]_{p \rightarrow p+1}=\frac{2\left[(p+1)^2+2\right]}{(p+1)\left[(p+1)^2+4\right]}\)
= \(\frac{2\left(p^2+2 p+3\right)}{(p+1)\left(p^2+2 p+5\right)}\)
84. Find \(L\left(e^t \cos ^2 t\right)\).
Solution:
L\(\left\{\cos ^2 t\right\}=L\left\{\frac{1+\cos 2 t}{2}\right\}=\frac{1}{2}\left[\frac{1}{p}+\frac{p}{p^2+4}\right]=\frac{2(p^2+2)}{p\left(p^2+4\right)}=f(p)\)
By First Shifting Theorem,\(L\left\{e^t \cos ^2 t\right\}=f(p-1)\)
= \(\left[\frac{2\left(p^2+2\right)}{p\left(p^2+4\right)}\right]_{p \rightarrow p-1}=\frac{2\left[(p-1)^2+2\right]}{(p-1)\left[(p-1)^2+4\right]}=\frac{2\left(p^2-2 p+3\right)}{(p1)\left(p^2-2 p+5\right)} \text {. }\)
85. Find \(L\left\{e^{3 t} \sin ^2 t\right\}\).
Solution:
We have \( L\left\{\sin ^2 t\right\}=L\left\{\frac{1-\cos 2 t}{2}\right\}=\frac{1}{2}[L\{1\}-L\{\cos 2 t\}]\)
= \(\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+4}\right]=\frac{2}{p\left(p^2+4\right)}=f(p), \text { say }\)
By First Shifting Theorem, \(L\left\{e^{3 t} \sin ^2 t\right\}=f(p-3)\)
= \(\left[\frac{2}{p\left(p^2+4\right)}\right]_{p \rightarrow p-3}=\frac{2}{(p-3)\left[(p-3)^2+4\right]}=\frac{2}{(p-3)\left(p^2-6 p+13\right)}\)
86. Find \(L\left(e^t \cos t \sin t\right)\).
Solution:
L\((\sin t \cos t)=L\left(\frac{\sin 2 t}{2}\right)=\frac{1}{2} L(\sin 2 t)=\frac{1}{2} \cdot \frac{2}{p^2+4}=\frac{1}{p^2+4}\)
L\(\left(e^t \cdot \sin t \cos t\right)=\left(\frac{1}{p^2+4}\right)_{p \rightarrow p-1}=\frac{1}{(p-1)^2+4}=\frac{1}{p^2+2 p+5} .\)
87. Find \(L\left\{e^t \sin 2 t \cos t\right\}\).
Solution:
L\(\{\sin 2 t \cos t\}=L\left\{\frac{\sin 3 t+\sin t}{2}\right\}=\frac{1}{2}[L\{\sin 3 t\}+L\{\sin t\}]=\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]\)
By First Shifting Theorem, \(L\left\{e^t \sin 2 t \cos t\right\}\)
= \(\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]_{p \rightarrow p-1}=\frac{1}{2}\left[\frac{3}{(p-1)^2+9}+\frac{1}{(p-1)^2+1}\right]\)
= \(\frac{1}{2}\left[\frac{3}{p^2-2 p+10}+\frac{1}{p^2-2 p+2}\right] .\)
88. Find \(L\left\{e^{4 t} \sin 2 t \cos t\right\}\).
Solution:
We have \(L\{\sin 2 t \cos t\}=L\left\{\frac{1}{2}(2 \sin 2 t \cos t)\right\}=L\left\{\frac{1}{2}(\sin 3 t+\sin t)\right\}\)
= \(\frac{1}{2}[L\{\sin 3 t\}+L\{\sin t\}]=\frac{1}{2}\left[\frac{3}{p^2+3^2}+\frac{1}{p^2+1^2}\right]\)
By First Shifting Theorem, \(L\left\{e^{4 t} \cdot \sin 2 t \cos t\right\}=\frac{1}{2}\left[\frac{3}{p^2+9}+\frac{1}{p^2+1}\right]_{p \rightarrow p-4}\)
= \(\frac{1}{2}\left[\frac{3}{(p-4)^2+9}+\frac{1}{(p-4)^2+1}\right]=\frac{1}{2}\left[\frac{3}{p^2-8 p+25}+\frac{1}{p^2-8 p+17}\right]\)
89. Find \(L\left[e^{-2 t} \sin 2 t \sin 3 t\right]\).
Solution:
L\([\sin 2 t \sin 3 t]=L\left[\frac{\cos t-\cos 5 t}{2}\right]=\frac{1}{2}[L(\cos t)-L(\cos 5 t)]\)
= \(\frac{1}{2}\left[\frac{p}{p^2+1}-\frac{p}{p^2+25}\right]=\frac{12 p}{\left(p^2+1\right)\left(p^2+25\right)}\)
L\(\left(e^{-2 t} \sin 2 t \sin 3 t\right)=\left[\frac{12 p}{\left(p^2+1\right)\left(p^2+25\right)}\right]_{p \rightarrow p+2}=\frac{12(p+2)}{\left\{(p+2)^2+1\right\}\left\{(p+2)^2+25\right\}}\)
= \(\frac{12 p+24}{\left(p^2+4 p+5\right)\left(p^2+4 p+29\right)} .\)
90. Find \(L\left\{e^{-t}(3 \sin 2 t-5 \cosh 2 t)\right\}\).
Solution:
L\(\langle 3 \sin 2 t-5 \cosh 2 t\}=3 L\{\sin 2 t\}-5 \dot{L}[\cosh 2 t]\)
= \(3 \frac{2}{p^2+2^2}-5 \frac{p}{p^2-2^2}=\frac{6}{p^2+4}-\frac{5 p}{p^2-4}\)
L\(\left\{e^{-t}(3 \sin 2 t-5 \cosh 2 t)\right\}=\left[\frac{6}{p^2+4}-\frac{5 p}{p^2-4}\right]_{p \rightarrow p+2}=\frac{6}{(p+1)^2+4}-\frac{5(p+1)}{(p+1)^2-4}\)
= \(\frac{6}{p^2+2 p+5}-\frac{5 p+5}{p^2+2 p-3}\)
91. Evaluate \(L\left\{e^t\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}\).
Solution:
L\(\left\{\cos 2 t+\frac{1}{2} \sinh 2 t\right\}=L\{\cos 2 t\}+\frac{1}{2} L[\sinh 2 t]=\frac{p}{p^2+2^2}+\frac{1}{2} \cdot \frac{2}{p^2-2^2}\)
= \(\frac{p}{p^2+4}+\frac{1}{p^2-4}=f(p), \text { say. }\) ∴ By First Shifting Theorem,
L\(\left\{e^t\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}=f(p-1)=\left(\frac{p}{p^2+4}+\frac{1}{p^2-4}\right)_{p \rightarrow p-1}\)
= \(\frac{p-1}{(p-1)^2+4}+\frac{1}{(p-1)^2-4}=\frac{p-1}{p^2-2 p+5}+\frac{1}{p^2-2 p-3} .\)
92. Evaluate \(L\left\{e^{-t}\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}\).
Solution:
L\(\left\{\cos 2 t+\frac{1}{2} \sinh 2 t\right\}=L\{\cos 2 t\}+\frac{1}{2} L[\sinh 2 t]=\frac{p}{p^2+2^2}+\frac{1}{2} \cdot \frac{2}{p^2-2^2}\)
+ \(\frac{p}{p^2+4}+\frac{1}{p^2-4}=f(p) \text {, say. }\) ∴ By First Shifting Theorem,
L\(\left\{e^{-t}\left(\cos 2 t+\frac{1}{2} \sinh 2 t\right)\right\}=f(p+1)=\left(\frac{p}{p^2+4}+\frac{1}{p^2-4}\right)_{p \rightarrow p+1}\)
= \(\frac{p+1}{(p+1)^2+4}+\frac{1}{(p+1)^2-4}=\frac{p+1}{p^2+2 p+5}+\frac{1}{p^2+2 p-3} .\)
93. Find Laplace transform of \(L\left\{e^{2 t}(3 \sinh 2 t-5 \cosh 2 t)\right\}\).
Solution:
By First Shifting Theorem,
L\(\left\{e^{2 t}(3 \sinh 2 t-5 \cosh 2 t)\right\}=[L\{3 \sinh 2 t-5 \cosh 2 t\}]_{p \rightarrow p-2}\)
= \(\left[\frac{6}{p^2-4}-\frac{5 p}{p^2-4}\right]_{p \rightarrow p-2}=\frac{6-5(p-2)}{(p-2)^2-4}=\frac{16-5 p}{p^2-4 p} .\)
94. Find L(sinh at cos at).
Solution:
L\((\sinh a t \cos a t)=L\left[\left(\frac{e^{a t}+e^{-a t}}{2}\right) \cos a t\right]=\frac{1}{2} L\left(e^{a t} \cos a t\right)+\frac{1}{2} L\left(e^{-a t} \cos a t\right)\)
L\((\cos a t)=\frac{p}{p^2+a^2} \Rightarrow L\left(e^{a t} \cos a t\right)=\left(\frac{p}{p^2+a^2}\right)_{p \rightarrow p-a}=\frac{p-a}{(p-a)^2+a^2}=\frac{p-a}{p^2-2 a p+2 a^2}\)
Similarly \(L\left(e^{-a t} \cos a t\right)=\frac{p+a}{p^2+2 a p+2 a^2}\)
∴ \(L(\sinh a t \cos a t)=\frac{1}{2}\left[\frac{p-a}{p^2-2 a p+2 a^2}\right]+\frac{1}{2}\left[\frac{p+a}{p^2+2 a p+2 a^2}\right]\)
95. Find L{sinh at sin at}.
Solution:
Sice \(\sinh a t \sin a t=\frac{1}{2}\left(e^{a t}-e^{-a t}\right) \sin a t \text {, we have }\)
L\(\{\sinh a t \sin a t\}=\frac{1}{2}\left[L\left\{e^{a t} \sin a t\right\}-L\left\{e^{-a t} \sin a t\right\}\right]\)
= \(\frac{1}{2}\left[\frac{a}{(p-a)^2+a^2}-\frac{a}{(p+a)^2+a^2}\right]=\frac{a}{2}\left[\frac{(p+a)^2+a^2-(p-a)^2-a^2}{\left\{(p-a)^2+a^2\right\}\left\{(p+a)^2+a^2\right\}}\right]\)
= \(\frac{a}{2} \frac{4 a p}{\left(p^2+2 a^2\right)^2-(2 a p)^2}=\frac{2 a^2 p}{p^4+4 a^4}\)
96. Find L{cosh at cos at}
Solution:
L\(\{\cosh a t \cos a t\}=\frac{1}{2} L\left\{\left(e^{a t}+e^{-a t}\right) \cos a t\right\}=\frac{1}{2}\left[L\left\{e^{a t} \cos a t\right\}+L\left\{e^{-a t} \cos a t\right\}\right]\)
= \(\frac{1}{2}\left[\frac{(p-a)}{(p-a)^2+a^2}+\frac{p+a}{(p+a)^2+a^2}\right]\), Using First Shifting Theorem
= \(\frac{1}{2}\left[\frac{(p-a)\left(p^2+2 a^2+2 a p\right)+(p+a)\left(p^2+2 a^2-2 a p\right)}{p^4+4 a^4}\right]=\frac{1}{2}\left[\frac{2 p\left(p^2+2 a^2\right)-4 p a^2}{p^4+4 a^4}\right]\)
= \(\frac{p\left(p^2+2 a^2-2 a^2\right)}{p^4+4 a^4}=\frac{p^3}{p^4+4 a^4}\)
97. Find L{cosh at sin bt}.
Solution:
L\(\{\cosh a t \sin b t\}=L\left\{\left(\frac{e^{a t}+e^{-a t}}{2}\right) \sin b t\right\}=\frac{1}{2}\left[L\left\{e^{a t} \sin b t\right\}+L\left\{e^{-a t} \sin b t\right\}\right]\)
By First Shifting Theorem,
L\(\{\cosh a t \sin b t\}=\frac{1}{2}\left[\{L(\sin b t)\}_{p \rightarrow p-a}+\{L(\sin b t)\}_{p \rightarrow p+a}\right]\)
= \(\frac{1}{2}\left[\left(\frac{b}{p^2+b^2}\right)_{p \rightarrow p-a}+\left(\frac{b}{p^2+b^2}\right)_{p \rightarrow p+a}\right]=\frac{1}{2}\left[\frac{b}{(p-a)^2+b^2}+\frac{b}{(p+a)^2+b^2}\right]\)
= \(\frac{b}{2}\left[\frac{1}{p^2-2 a p+a^2+b^2}+\frac{1}{p^2+2 a p+a^2+b^2}\right]=\frac{b}{2}\left[\frac{\dot{2}\left(p^2+a^2+b^2\right)}{\left[(p-a)^2+b^2\right]\left[(p+a)^2+b^2\right]}\right]\)
= \(\frac{b\left(p^2+a^2+b^2\right)}{\left[(p-a)^2+b^2\right]\left[(p+a)^2+b^2\right]}\)
98. Evaluate \(L\left\{t^2 e^{-2 t}\right\}\).
Solution:
We know that \(L\left(t^2\right)=\frac{2 !}{t^3}=\frac{2}{p^3} \text { then } L\left(e^{-2 t} t^2\right)=\left(\frac{2}{p^3}\right)_{p \rightarrow p+2}=\frac{2}{(p+2)^3} \text {. }\)
99. Find \(L\left(e^{-4 t} t^2\right)\).
Solution:
We know that \(L\left(t^2\right)=\frac{2 !}{t^3}=\frac{2}{p^3} \text { then } L\left(e^{-4 t} t^2\right)=\left(\frac{2}{p^3}\right)_{p \rightarrow p+4}=\frac{2}{(p+4)^3}\)
100. Find \(\dot{L}\left\{e^{-3 t}\left(t^2+1\right)\right\}\).
Solution:
L\(\left\{\left(t^2+1\right)\right\}=L\left\{t^2\right\}+L\{1\}=\frac{2 !}{p^3}+\frac{1}{p}\)
L\(\left\{e^{-3 t}\left(t^2+1\right)\right\}=\left(\frac{2}{p^3}+\frac{1}{p}\right)_{p \rightarrow p+3}=\frac{2}{(p+3)^3}+\frac{1}{p+3}=\frac{2+(p+3)^2}{(p+3)^3}=\frac{p^2+6 p+11}{(p+3)^3}\)
101. Find \(L\left\{(t+3)^2 e^t\right\}\).
Solution:
We have \(L\left\{(t+3)^2\right\}=L\left\{t^2+6 t+9\right\}=\frac{2}{p^3}+\frac{6}{p^2}+\frac{9}{p}=f(p)\)
By First Shifting Theorem, \(L\left\{e^t(t+3)^2\right\}=f(p-1)=\left[\frac{2}{p^3}+\frac{6}{p^2}+\frac{9}{p}\right]_{p \rightarrow p-1}\)
= \(\frac{2}{(p-1)^3}+\frac{6}{(p-1)^2}+\frac{9}{(p-1)}=\frac{9 p^2-12 p+5}{(p-1)^3}\)
102. Find \(L\left\{(t+2)^2 e^t\right\}\).
Solution:
By First Shifting Theorem,
L\(\left\{(t+2)^2 e^t\right\}=\left[L\left\{(t+2)^2\right\}\right]_{p \rightarrow p-1}=\left[L\left\{t^2+4 t+4\right\}\right]_{p \rightarrow p-1}\)
= \(\left[\frac{2}{p^3}+\frac{4}{p^2}+\frac{4}{p}\right]_{p \rightarrow p-1}=\left[\frac{2+4 p+4 p^2}{p^3}\right]_{p \rightarrow p-1}=\frac{2+4(p-1)+4(p-1)^2}{(p-1)^3}\)
= \(\frac{2\left[1+2 p-2+2 p^2-4 p+2\right]}{(p-1)^3}=\frac{2\left(2 p^2-2 p+1\right)}{(p-1)^3}\)
103. Find \(L\left\{\left(1+t e^{-t}\right)^3\right\}\).
Solution:
We have \(\left(1+t e^{-t}\right)^3=1+3 t e^{-t}+3 t^2 e^{-2 t}+t^3 e^{-3 t} \text {. }\)
∴ \(L\left\{\left(1+t e^{-t}\right)^3\right\}=L\left\{1+3 t e^{-t}+3 t^2 e^{-2 t}+t^3 e^{-3 t}\right\}\)
= \(L\{1\}+3 L\left\{t e^{-t}\right\}+3 L\left\{t^2 e^{-2 t}\right\}+L\left\{t^3 e^{-3 t}\right\}, \text { using Linearity Property }\)
= \(\frac{1}{p}+3 \cdot \frac{1}{(p+1)^2}+3 \cdot \frac{2 !}{(p+2)^2}+\frac{3 !}{(p+3)^4}=\frac{1}{p}+\frac{3}{(p+1)^2}+\frac{6}{(p+2)^3}+\frac{6}{(p+3)^4}\)
104. Show that
1) \(L\{t \sin a t\}=\frac{2 a p}{\left(p^2+a^2\right)^2}\)
2) \(L\{t \cos a t\}=\frac{p^2-a^2}{\left(p^2+a^2\right)^2}\)
Solution:
Since, \(L\{t\}=\frac{1}{p^2} \text {, we have } L\left\{t e^{i a t}\right\}=\frac{1}{(p-i a)^2}=\frac{(p+i a)^2}{(p-i a)^2(p+i a)^2}\)
Equating the real and imaginary parts on both sides, we obtain
⇒ \(L\{t(\cos a t+i \sin a t)\}=\frac{\left(p^2-\dot{a}^2\right)+i 2 a p}{\left(p^2+a^2\right)^2}\)
1) \(L\{t \sin a t\}=\frac{2 a p}{\left(p^2+a^2\right)^2}\)
2) \(L\{t \cos a t\}=\frac{p^2-a^2}{\left(p^2+a^2\right)^2}\)
105. Find the Laplace transform of \(t e^{2 t}\) sin3t.
Solution:
Since \(L\{t\}=\frac{1}{p^2}\), we have \(L\left\{t e^{i 3 t}\right\}=\frac{1}{(p-3 i)^2}=\frac{(p+3 i)^2}{(p-3 i)^2(p+3 i)^2}\)
⇒ \(L\{t \cos 3 t+i t \sin 3 t\}=\frac{\left(p^2-9\right)+i 6 p}{\left(p^2+9\right)^2}\)
Equating imaginary parts on both sides, we have \(L\{t \sin 3 t\}=\frac{6 p}{\left(p^2+9\right)^2}\)
Now applying the First Shifting theorem, we have \(L\left\{e^{2 t} t \sin 3 t\right\}=[L\{t \sin 3 t\}]_{p \rightarrow p-2}=\left[\frac{6 p}{\left(p^2+9\right)^2}\right]_{p \rightarrow p-2}=\frac{6(p-2)}{\left(p^2-4 p+13\right)^2}\)
106. State and prove the second translation or shifting theorem.
Solution:
Statement: If L[F(t)]=f(p) and \(G(t)=\left\{\begin{array}{l}F(t-a), t>a \text {. } \\ 0, t<a\end{array}\right.\) then \(L[G(t)]=e^{-a p} f(p)\).
Proof: \(L(G(t)]=\int_0^{\infty} e^{-s t} G(t) d t=\int_0^a e^{-p t} G(t) d t+\int_a^{\infty} e^{-p t} G(t) d t\)
= \(\int_0^a e^{-p t} 0 d t+\int_a^{\infty} e^{-p t} F(t-a) d t=\int_a^{\infty} e^{-p t} F(t-a) d t\)
Put u=t-a. Then, t=u+a and dt=du when t=a, u=0 and \(t \rightarrow \infty, u \rightarrow \infty\)
∴ \(L[G(t)]=\int_0^{\infty} e^{-p(u+a)} F(u) d u=e^{-a p} \int_0^{\infty} e^{-p u} F(u) d u=e^{-a p} f(p)\)
107. If L[F(t)]=f(p) and a>0, then prove that \(L\{F(t-a) H(t-a)\}=e^{-a p} f(p)\), where \(H(t)= \begin{cases}1, & \text { if } t>0 \\ 0, & \text { if } t<0\end{cases}\)
Solution:
By definition, \(L\{F(t-a) H(t-a)\}=\int_0^{\infty} e^{-p t} F(t-a) H(t-a) d t \rightarrow(1)\)
Put t-a=u so that dt=du.
Also when t=0, u=-a when \(t \rightarrow \infty, u \rightarrow \infty\).
Then \(L\{F(t-a) H(t-a)\}=\int_0^{\infty} e^{-p(u+a)} F(u) H(u) d u\) by (1)
⇒ \(\int_{-a}^0 e^{-p(u+a)} F(u) H(u) d u+\int_0^{\infty} e^{-p(u+a)} F(u) H(u) d u\)
= \(\int_{-a}^0 e^i-p(u+a) F(u) \cdot 0 d u+\int_0^{\infty} e^{-p(u+a)} F(u) \cdot 1 d u\), by Definition of H(t)
= \(\int_{-a}^0 e^{-p(u+a)} F(u) d u=e^{-p a} \int_0^{\infty} e^{-p u} F(u) d u=e^{-p a} \int_0^{\infty} e^{-p t} F(t) d t,\)
= \(e^{-p a} L\{F(t)\}=e^{-a p} f(p)\)
108. Find L[G(t)], where \(G(t)= \begin{cases}e^{t-a}, & t>a \\ 0, & t<a\end{cases}\).
Solution:
Let \(F(t)=e^t \text {. Then, } L[F(t)]=f(p)=\frac{1}{p-1}\)
∴ \(G(t)=\left\{\begin{array}{l}
F(t-a), t>a. \\
0, \quad t<a
\end{array}\right.\)
Then \(L[G(t)]=e^{-a p} f(p)=e^{-a p} \frac{1}{p-1}=\frac{e^{-a p}}{p-1}, p>1\)
109. Find L{G(t)} where \(G(t)=\left\{\begin{array}{cc}
\sin (t-2 \pi / 3), & t>2 \pi / 3 \\
0, & t<2 \pi / 3
\end{array}\right.\)
Solution:
Let \(F(t)=\sin t \text { so that } G(t)=\left\{\begin{array}{cl}
F(t-2 \pi / 3), & t>2 \pi / 3 . \\
0, & t<2 \pi / 3
\end{array}\right.\)
We know that, \(L(F(t)]=L(\sin t)=\frac{1}{p^2+1}=f(p)\)
∴ \(L[G(t)]=e^{\frac{-2 \pi}{3} p} \frac{1}{p^2+1}=\frac{e^{-2 \pi p / 3}}{p^2+1}, p>0\)
110. Find L{F(t)} where \(F(t)=\left\{\begin{array}{rr}
\sin \left(t-\frac{\pi}{3}\right), & t>\frac{\pi}{3} \\
0, & t<\frac{\pi}{3}
\end{array}\right\}\).
Solution:
Let \(G(t)=\sin t \text {. }\)
∴ \(L\{G(t)\}=L\{\sin t\}=\frac{1}{p^2+1}=f(p)\)
Now, \(F(t)=\left\{\begin{aligned}
\sin (t-\pi / 3), & t>\pi / 3 \\
0, & t>\pi / 3
\end{aligned}\right.\)
Applying second shifting theorem \(L\{G(t)\}=e^{-\pi p / 3} f(p)=\frac{e^{-\pi p / 3}}{p^2+1}\)
111. Find L[G(t)], where \(G(t)=\left\{\begin{array}{cc}
\sin 2 t, & 0<t<\pi \\
0, & t>\pi
\end{array}\right.\).
Solution:
L\([G(t)]=\int_0^{\infty} e^{-p t} G(t) d t 2=\int_0^\pi e^{-p t} \sin 2 t d t+\int_\pi^{\infty} e^{-p t}(0) d t\)
= \(\left[e^{-p t} \frac{-(p \sin 2 t+2 \cos 2 t)}{p^2+4}\right]_0^\pi=0\)
∴ \(\left[ \int e^{a x} \sin b x d x=\frac{e^{a x}}{a^2+b^2}(a \sin b x-b \cos b x)\right]\)
112. Find L{G(t)} where \(G(t)=\left\{\begin{array}{cl}
\cos (-t-\pi / 3), & \text { if } t>\pi / 3 \\
0, & \text { if } t<\pi / 3
\end{array}\right.\)
Solution:
Let \(F(t)=\cos t \text {. }\)
∴ \(L\{F(t)\}=L\{\cos t\}=\frac{p}{p^2+1}=f(p)\)
Now \(G(t)=\left\{\begin{array}{cc}
F(t-\pi / 3)=\cos (t-\pi / 3), & t>\pi / 3 \\
0, & t<\pi / 3
\end{array}\right.\)
Applying the second Shifting theorem, we get \(L\{(t)\}=e^{-\pi p / 3}\left(\frac{p}{p^2+1}\right)=\frac{p e^{-\pi p / 3}}{p^2+1}\)
Proceeding as above, we get L\(L[G(t)]=e^{-2 \pi p / p}\left(\frac{1}{p^2+1}\right)=\frac{p e^{-2 \pi p / 3}}{p^2+1}\)
113. Find L{F(t)} where \(F(t)=\left\{\begin{array}{cl}
\cos (t-2 \pi / 3), & \text { if } t>2 \pi / 3 \\
0, & \text { if } t<2 \pi / 3
\end{array}\right.\)
Solution:
Let \(G(t)=\cos t \text {. }\)
∴ \(L\{G(t)\}=L\{\cos t\}=\frac{p}{p^2+1}=f(p)\)
Now, \(F(t)=\left\{\begin{array}{cc}
G(t-2 \pi / 3)=\cos (t-2 \pi / 3), & t>2 \pi / 3 \\
0, & t<2 \pi / 3
\end{array}\right.\)
Applying second shifting theorem\(L\{F(t)\}=e^{-2 \pi p / 3}\left(\frac{p}{p^2+1}\right)=\frac{p e^{-2 \pi p / 3}}{p^2+1}\)
114. Find the Laplace transform of \((t-2)^3 u(t-2)\).
Solution:
Comparing the given function with \(f(t-a) u(t-a) \text {, we have } a=2 \text { and } f(t)=t^3\)
∴ \(L\{F(t)\}=L\left\{t^3\right\}=\frac{3 !}{p^4}=\frac{6}{p^4}=f(p)\)
Applying second shifting theorem \(L\left\{(t-2)^3 u(t-2)\right\}=e^{-2 p} \frac{6}{p^4}=\frac{6 e^{-2 p}}{p^4}\)
115. Find the Laplace transform of \(e^{-3 t} u(t-2)\).
Solution:
L\(\left\{e^{-3 t} u(t-2)\right\}=L\left\{e^{-3(t-2)} e^{-6} u(t-2)\right\}=e^{-6} L\left\{e^{-3(t-2)} u(t-2)\right\}\)
Taking,\(F(t)=e^{-3 t}, f(p)=\frac{1}{p+3}\)and using second shifting theorem, we have
L\(\left\{e^{-3 t} u(t-2)\right\} = e^{-6} e^{-2 p} \frac{1}{p+3}=\frac{e^{-2(p+3)}}{p+3}\)
116. State and prove a change of scale property.
Solution:
Statement: If \(L[F(t)]=f(p), \text { then } L[F(a t)]=\frac{1}{a} f\left(\frac{p}{a}\right)\)
Proof: We know that \(L[F(a t)]=\int_0^{\infty} e^{-p t} F(a t) d t\)
Put u = at Then, t = \(\frac{u}{a}\) and \(d t=\frac{d u}{a}\)
∴ \(L[F(a t)]=\int_0^{\infty} e^{-p u / a} F(u) \frac{d u}{a}=\frac{1}{a} \int_0^{\infty} e^{-p W a} F(u) d u=\frac{1}{a} \int_0^{\infty} e^{-(b / a) u} F(u) d u=\frac{1}{a} f\left(\frac{p}{a}\right)\)
117. Find L(cos 4t) by using the change of scale property.
Solution:
We have that \(L(\cos t)=\frac{p}{p^2+1}=f(p)\)
Then \(L(\cos 4 t)=\frac{1}{4} f\left(\frac{p}{4}\right)=\frac{1}{4} \cdot \frac{p / 4}{(p / 4)^2+1}=\frac{p}{p^2+16}, p>0 \text {. }\)
118. Find L{cos 5t} by using the change of scale property.Solution:
We have that \(L(\cos t)=\frac{p}{p^2+1}=f(p)\)
Then \(L(\cos 5 t)=\frac{1}{5} f\left(\frac{p}{5}\right)=\frac{1}{5} \cdot \frac{p / 5}{(p / 5)^2+1}=\frac{p}{p^2+25}, p>0 \text {. }\)
119. Find L[sinh 2t] by using the change of scale property.
Solution:
We have \(L[\sinh t]=\frac{1}{p^2-1}=f(p)\)
∴ \(L[\sinh 2 t]=\frac{1}{2} \cdot f\left(\frac{p}{2}\right)=\frac{1}{2} \cdot \frac{1}{(p / 2)^2-1}=\frac{2}{p^2-4}\)
120. Find \(L\left\{\sin ^2(a t)\right\}\) by using change of scale property.
Solution:
Since \(\sin ^2(a t)=\frac{1-\cos 2 a t}{2} \text {, we have }\)
L\(\left\{\sin ^2(a t)\right\}=\frac{1}{2} L\{(1-\cos 2 a t)\}=\frac{1}{2}[L(1)-L\{\cos 2 a t\}]=\frac{1}{2} \cdot \frac{1}{p}-\frac{1}{2} L\{\cos 2 a t\}\)
Taking \(F(t)=\cos t, f(p)=\frac{p}{p^2+1}\)
∴ \(L\{\cos (2 a t)\}=\frac{1}{2 a} \frac{p / 2 a}{(P / 2 a)^2+1}=\frac{p}{p^2+4 a^2}\)
Hence \(L\left\{\sin ^2 a t\right\}=\frac{1}{2 p}-\frac{1}{2} \frac{p}{p^2+4 a^2}=\frac{1}{2}\left[\frac{1}{p}-\frac{p}{p^2+4 a^2}\right]=\frac{2 a^2}{p\left(p^2+4 a^2\right)} \text {. }\)
121. If \(L\{F(t)\}=\frac{1}{p} e^{-1 / p}\), prove that \(L\left\{e^{-t} F(3 t)\right\}=\frac{e^{-3 /(p+1)}}{p+1}\)
Solution:
L\(\{F(t)\}=\frac{1}{p} e^{-1 / p}=f(p)\)
∴ By the Change of Scale property, we have \(F(3 t)=\frac{1}{3} f\left(\frac{p}{3}\right)=\frac{1}{3} \cdot \frac{3}{p} e^{-3 / p}=\frac{1}{p} e^{-3 / p}\)
Now applying the First Shifting theorem, we get \(L\left\{e^{-t} F(3 t)\right\}=\frac{e^{-3 /(p+1)}}{p+1}\).
122. If \(L[F(t)]=\frac{9 p^2-12 p+15}{(p-1)^3}\) using change of scale property
Solution:
Given \(L[F(t)]=\frac{9 p^2-12 p+15}{(p-1)^3}=f(p)\). By Change of Scale Property,
L\([F(3 t)]=\frac{1}{3} f\left(\frac{p}{3}\right)=\frac{1}{3} \cdot \frac{9(p / 3)^2-12(p / 3)+15}{(p / 3-1)^3}=\frac{9\left(p^2-4 p+15\right)}{(p-3)^3}\)
123. If \(L\{F(t)\}=\frac{p^2-p+1}{(2 p+1)^2(p-1)}\) then show that \(L\{F(2 t)\}=\frac{p^2-2 p+4}{4(p+1)^2(p-2)}\) by applying change of scale property,
Solution:
Given \(L\{F(t)\}=\frac{p^2-p+1}{(2 p+1)^2(p-1)}=f(p)\)
By Change of scale property,
L\(\{F(2 t)\}=\frac{1}{2} f\left(\frac{p}{2}\right)=\frac{1}{2}\left[\frac{\left(\frac{p}{2}\right)^2-\frac{p}{2}+1}{\left(2 \cdot \frac{p}{2}+1\right)\left(\frac{p}{2}-1\right)}\right]=\frac{1}{2}\left[\frac{-\frac{p^2-2 p+4}{4}}{\frac{(p+1)^2(p-2)}{4}}\right]=\frac{1}{4} \frac{p^2-2 p+4}{(p+1)^2(p-2)}\)