Here are the solutions to problems 11, 12, 13, 14, and 15:
11. Find L[cos3t] and L[sin3t]
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For L[cos3t]:
Step 1: Use the trigonometric identity cos(3t)=4cos3t−3cost.
Rearrange to express cos3t:
cos3t=41(cos(3t)+3cost)
Step 2: Apply the Laplace transform using linearity.
L[cos3t]=L[41(cos(3t)+3cost)]
L[cos3t]=41(L[cos(3t)]+3L[cost])
Step 3: Use the standard Laplace transform formula L[cos(at)]=s2+a2s.
L[cos(3t)]=s2+32s=s2+9s
L[cost]=s2+12s=s2+1s
Step 4: Substitute and simplify.
L[cos3t]=41(s2+9s+3s2+1s)
L[cos3t]=4s(s2+91+s2+13)
L[cos3t]=4s((s2+9)(s2+1)s2+1+3(s2+9))
L[cos3t]=4s((s2+9)(s2+1)s2+1+3s2+27)
L[cos3t]=4s((s2+9)(s2+1)4s2+28)
L[cos3t]=(s2+9)(s2+1)s(s2+7)
L[\cos^3 t] = \frac{s(s^2+7){(s^2+1)(s^2+9)}}
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For L[sin3t]:
Step 1: Use the trigonometric identity sin(3t)=3sint−4sin3t.
Rearrange to express sin3t:
sin3t=41(3sint−sin(3t))
Step 2: Apply the Laplace transform using linearity.
L[sin3t]=L[41(3sint−sin(3t))]
L[sin3t]=41(3L[sint]−L[sin(3t)])
Step 3: Use the standard Laplace transform formula L[sin(at)]=s2+a2a.
L[sint]=s2+121=s2+11
L[sin(3t)]=s2+323=s2+93
Step 4: Substitute and simplify.
L[sin3t]=41(3s2+11−s2+93)
L[sin3t]=43(s2+11−s2+91)
L[sin3t]=43((s2+1)(s2+9)(s2+9)−(s2+1))
L[sin3t]=43((s2+1)(s2+9)s2+9−s2−1)
L[sin3t]=43((s2+1)(s2+9)8)
L[sin3t]=(s2+1)(s2+9)6
L[\sin^3 t] = \frac{6{(s^2+1)(s^2+9)}}
12. Find L−1[s2−a2a]
Step 1: This is a direct application of a standard inverse Laplace transform formula.
Recall the Laplace transform of the hyperbolic sine function:
L[sinh(at)]=s2−a2a
Step 2: Therefore, the inverse Laplace transform is:
L−1[s2−a2a]=sinh(at)
13. Find L−1[(s−2)2+91]
Step 1: Identify the form of the expression. It resembles L−1[(s−a)2+b2b].
Here, a=2 and b2=9, so b=3.
Step 2: Adjust the numerator to match the required form for sin(bt).
L−1[(s−2)2+321]=31L−1[(s−2)2+323]
Step 3: Apply the inverse Laplace transform using the shifting property L−1[F(s−a)]=eatf(t).
We know L−1[s2+b2b]=sin(bt).
So, L−1[(s−2)2+323]=e2tsin(3t).
Step 4: Combine the results.
L−1[(s−2)2+91]=31e2tsin(3t)
L^{-1}\left[\frac{1}{(s-2)^2+9}\right] = \frac{1{3} e^{2t}\sin(3t)}
14. Find L[sin3tsin5t]
Step 1: Use the product-to-sum trigonometric identity: sinAsinB=21[cos(A−B)−cos(A+B)].
Let A=5t and B=3t.
sin5tsin3t=21[cos(5t−3t)−cos(5t+3t)]
sin5tsin3t=21[cos(2t)−cos(8t)]
Step 2: Apply the Laplace transform using linearity.
L[sin3tsin5t]=L[21(cos(2t)−cos(8t))]
L[sin3tsin5t]=21(L[cos(2t)]−L[cos(8t)])
Step 3: Use the standard Laplace transform formula L[cos(at)]=s2+a2s.
L[cos(2t)]=s2+22s=s2+4s
L[cos(8t)]=s2+82s=s2+64s
Step 4: Substitute and simplify.
L[sin3tsin5t]=21(s2+4s−s2+64s)
L[sin3tsin5t]=2s((s2+4)(s2+64)(s2+64)−(s2+4))
L[sin3tsin5t]=2s((s2+4)(s2+64)s2+64−s2−4)
L[sin3tsin5t]=2s((s2+4)(s2+64)60)
L[sin3tsin5t]=(s2+4)(s2+64)30s
L[\sin 3t \sin 5t] = \frac{30s{(s^2+4)(s^2+64)}}
15. Find L−1[(s2+6s+13)2s+3]
Step 1: Complete the square in the denominator.
s2+6s+13=(s2+6s+9)+4=(s+3)2+22
Step 2: Rewrite the expression using the completed square.
((s+3)2+22)2s+3
Step 3: This expression is of the form F(s−a) where a=−3. Let G(s)=(s2+22)2s.
Then the given expression is G(s−(−3)).
We need to find L−1[G(s)]=L−1[(s2+22)2s].
From , we know that L−1[(s2+a2)2s]=2atsin(at).
For a=2:
L−1[(s2+22)2s]=2(2)tsin(2t)=4tsin(2t)
Step 4: Apply the shifting property L−1[F(s−a)]=eatf(t).
Here, a=−3 and f(t)=4tsin(2t).
L−1[((s+3)2+22)2s+3]=e−3t(4tsin(2t))
L^{-1}\left[\frac{s+3}{(s^2+6s+13)^2}\right] = \frac{t e^{-3t \sin(2t)}{4}}
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