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Find L[3 t] and L[3 t]

Mathematics

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Here are the solutions to problems 11, 12, 13, 14, and 15:

11. Find L[cos⁡3t]L[\cos^3 t] and L[sin⁡3t]L[\sin^3 t]

  • For L[cos⁡3t]L[\cos^3 t]:

  1. Use the trigonometric identity cos⁡(3t)=4cos⁡3t−3cos⁡t\cos(3t) = 4\cos^3 t - 3\cos t.

    Rearrange to express $\cos^3 t$:
$$
\cos^3 t = \frac{1}{4}(\cos(3t) + 3\cos t)
$$

  2. Apply the Laplace transform using linearity.

    $$
L[\cos^3 t] = L\left[\frac{1}{4}(\cos(3t) + 3\cos t)\right]
$$
$$
L[\cos^3 t] = \frac{1}{4}(L[\cos(3t)] + 3L[\cos t])
$$

  3. Use the standard Laplace transform formula L[cos⁡(at)]=ss2+a2L[\cos(at)] = \frac{s}{s^2+a^2}.

    $$
L[\cos(3t)] = \frac{s}{s^2+3^2} = \frac{s}{s^2+9}
$$
$$
L[\cos t] = \frac{s}{s^2+1^2} = \frac{s}{s^2+1}
$$

  4. Substitute and simplify.

    $$
L[\cos^3 t] = \frac{1}{4}\left(\frac{s}{s^2+9} + 3\frac{s}{s^2+1}\right)
$$
$$
L[\cos^3 t] = \frac{s}{4}\left(\frac{1}{s^2+9} + \frac{3}{s^2+1}\right)
$$
$$
L[\cos^3 t] = \frac{s}{4}\left(\frac{s^2+1 + 3(s^2+9)}{(s^2+9)(s^2+1)}\right)
$$
$$
L[\cos^3 t] = \frac{s}{4}\left(\frac{s^2+1 + 3s^2+27}{(s^2+9)(s^2+1)}\right)
$$
$$
L[\cos^3 t] = \frac{s}{4}\left(\frac{4s^2+28}{(s^2+9)(s^2+1)}\right)
$$
$$
L[\cos^3 t] = \frac{s(s^2+7)}{(s^2+9)(s^2+1)}
$$
$$
L[\cos^3 t] = \boxed{\frac{s(s^2+7)}{(s^2+1)(s^2+9)}}
$$
    • For L[sin⁡3t]L[\sin^3 t]:

  5. Use the trigonometric identity sin⁡(3t)=3sin⁡t−4sin⁡3t\sin(3t) = 3\sin t - 4\sin^3 t.

    Rearrange to express $\sin^3 t$:
$$
\sin^3 t = \frac{1}{4}(3\sin t - \sin(3t))
$$

  6. Apply the Laplace transform using linearity.

    $$
L[\sin^3 t] = L\left[\frac{1}{4}(3\sin t - \sin(3t))\right]
$$
$$
L[\sin^3 t] = \frac{1}{4}(3L[\sin t] - L[\sin(3t)])
$$

  7. Use the standard Laplace transform formula L[sin⁡(at)]=as2+a2L[\sin(at)] = \frac{a}{s^2+a^2}.

    $$
L[\sin t] = \frac{1}{s^2+1^2} = \frac{1}{s^2+1}
$$
$$
L[\sin(3t)] = \frac{3}{s^2+3^2} = \frac{3}{s^2+9}
$$

  8. Substitute and simplify.

    $$
L[\sin^3 t] = \frac{1}{4}\left(3\frac{1}{s^2+1} - \frac{3}{s^2+9}\right)
$$
$$
L[\sin^3 t] = \frac{3}{4}\left(\frac{1}{s^2+1} - \frac{1}{s^2+9}\right)
$$
$$
L[\sin^3 t] = \frac{3}{4}\left(\frac{(s^2+9) - (s^2+1)}{(s^2+1)(s^2+9)}\right)
$$
$$
L[\sin^3 t] = \frac{3}{4}\left(\frac{s^2+9 - s^2-1}{(s^2+1)(s^2+9)}\right)
$$
$$
L[\sin^3 t] = \frac{3}{4}\left(\frac{8}{(s^2+1)(s^2+9)}\right)
$$
$$
L[\sin^3 t] = \frac{6}{(s^2+1)(s^2+9)}
$$
$$
L[\sin^3 t] = \boxed{\frac{6}{(s^2+1)(s^2+9)}}
$$

    12. Find L−1[as2−a2]L^{-1}\left[\frac{a}{s^2-a^2}\right]

  9. This is a direct application of a standard inverse Laplace transform formula.

    Recall the Laplace transform of the hyperbolic sine function:

    L[sinh⁡(at)]=as2−a2L[\sinh(at)] = \frac{a}{s^2-a^2}

  10. Therefore, the inverse Laplace transform is

    L−1[as2−a2]=sinh⁡(at)L^{-1}\left[\frac{a}{s^2-a^2}\right] = \boxed{\sinh(at)}

    13. Find L−1[1(s−2)2+9]L^{-1}\left[\frac{1}{(s-2)^2+9}\right]

  11. Identify the form of the expression.

    It resembles L−1[b(s−a)2+b2]L^{-1}\left[\frac{b}{(s-a)^2+b^2}\right].
    Here, a=2a=2 and b2=9b^2=9, so b=3b=3.

  12. Adjust the numerator to match the required form for sin⁡(bt)\sin(bt).

    L−1[1(s−2)2+32]=13L−1[3(s−2)2+32]L^{-1}\left[\frac{1}{(s-2)^2+3^2}\right] = \frac{1}{3} L^{-1}\left[\frac{3}{(s-2)^2+3^2}\right]

  13. Apply the inverse Laplace transform using the shifting property L−1[F(s−a)]=eatf(t)L^{-1}[F(s-a)] = e^{at}f(t).

    We know L−1[bs2+b2]=sin⁡(bt)L^{-1}\left[\frac{b}{s^2+b^2}\right] = \sin(bt).
    So, L−1[3(s−2)2+32]=e2tsin⁡(3t)L^{-1}\left[\frac{3}{(s-2)^2+3^2}\right] = e^{2t}\sin(3t).

  14. Combine the results.

    L−1[1(s−2)2+9]=13e2tsin⁡(3t)L^{-1}\left[\frac{1}{(s-2)^2+9}\right] = \frac{1}{3} e^{2t}\sin(3t) L−1[1(s−2)2+9]=13e2tsin⁡(3t)L^{-1}\left[\frac{1}{(s-2)^2+9}\right] = \boxed{\frac{1}{3} e^{2t}\sin(3t)}

    14. Find L[sin⁡3tsin⁡5t]L[\sin 3t \sin 5t]

  15. Use the product-to-sum trigonometric identity

    sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)].
    Let A=5tA=5t and B=3tB=3t.

    sin⁡5tsin⁡3t=12[cos⁡(5t−3t)−cos⁡(5t+3t)]\sin 5t \sin 3t = \frac{1}{2}[\cos(5t-3t) - \cos(5t+3t)] sin⁡5tsin⁡3t=12[cos⁡(2t)−cos⁡(8t)]\sin 5t \sin 3t = \frac{1}{2}[\cos(2t) - \cos(8t)]

  16. Apply the Laplace transform using linearity.

    L[sin⁡3tsin⁡5t]=L[12(cos⁡(2t)−cos⁡(8t))]L[\sin 3t \sin 5t] = L\left[\frac{1}{2}(\cos(2t) - \cos(8t))\right] L[sin⁡3tsin⁡5t]=12(L[cos⁡(2t)]−L[cos⁡(8t)])L[\sin 3t \sin 5t] = \frac{1}{2}(L[\cos(2t)] - L[\cos(8t)])

  17. Use the standard Laplace transform formula L[cos⁡(at)]=ss2+a2L[\cos(at)] = \frac{s}{s^2+a^2}.

    L[cos⁡(2t)]=ss2+22=ss2+4L[\cos(2t)] = \frac{s}{s^2+2^2} = \frac{s}{s^2+4} L[cos⁡(8t)]=ss2+82=ss2+64L[\cos(8t)] = \frac{s}{s^2+8^2} = \frac{s}{s^2+64}

  18. Substitute and simplify.

    L[sin⁡3tsin⁡5t]=12(ss2+4−ss2+64)L[\sin 3t \sin 5t] = \frac{1}{2}\left(\frac{s}{s^2+4} - \frac{s}{s^2+64}\right) L[sin⁡3tsin⁡5t]=s2((s2+64)−(s2+4)(s2+4)(s2+64))L[\sin 3t \sin 5t] = \frac{s}{2}\left(\frac{(s^2+64) - (s^2+4)}{(s^2+4)(s^2+64)}\right) L[sin⁡3tsin⁡5t]=s2(s2+64−s2−4(s2+4)(s2+64))L[\sin 3t \sin 5t] = \frac{s}{2}\left(\frac{s^2+64 - s^2-4}{(s^2+4)(s^2+64)}\right) L[sin⁡3tsin⁡5t]=s2(60(s2+4)(s2+64))L[\sin 3t \sin 5t] = \frac{s}{2}\left(\frac{60}{(s^2+4)(s^2+64)}\right) L[sin⁡3tsin⁡5t]=30s(s2+4)(s2+64)L[\sin 3t \sin 5t] = \frac{30s}{(s^2+4)(s^2+64)} L[sin⁡3tsin⁡5t]=30s(s2+4)(s2+64)L[\sin 3t \sin 5t] = \boxed{\frac{30s}{(s^2+4)(s^2+64)}}

    15. Find L−1[s+3(s2+6s+13)2]L^{-1}\left[\frac{s+3}{(s^2+6s+13)^2}\right]

  19. Complete the square in the denominator.

    s2+6s+13=(s2+6s+9)+4=(s+3)2+22s^2+6s+13 = (s^2+6s+9) + 4 = (s+3)^2+2^2

  20. Rewrite the expression using the completed square.

    s+3((s+3)2+22)2\frac{s+3}{((s+3)^2+2^2)^2}

  21. This expression is of the form F(s−a)F(s-a) where a=−3a=-3.

    Let G(s)=s(s2+22)2G(s) = \frac{s}{(s^2+2^2)^2}.
    Then the given expression is G(s−(−3))G(s-(-3)).
    We need to find L−1[G(s)]=L−1[s(s2+22)2]L^{-1}[G(s)] = L^{-1}\left[\frac{s}{(s^2+2^2)^2}\right].
    From , we know that L−1[s(s2+a2)2]=tsin⁡(at)2aL^{-1}\left[\frac{s}{(s^2+a^2)^2}\right] = \frac{t \sin(at)}{2a}.
    For a=2a=2:

    L−1[s(s2+22)2]=tsin⁡(2t)2(2)=tsin⁡(2t)4L^{-1}\left[\frac{s}{(s^2+2^2)^2}\right] = \frac{t \sin(2t)}{2(2)} = \frac{t \sin(2t)}{4}

  22. Apply the shifting property L−1[F(s−a)]=eatf(t)L^{-1}[F(s-a)] = e^{at}f(t).

    Here, a=−3a=-3 and f(t)=tsin⁡(2t)4f(t) = \frac{t \sin(2t)}{4}.

    L−1[s+3((s+3)2+22)2]=e−3t(tsin⁡(2t)4)L^{-1}\left[\frac{s+3}{((s+3)^2+2^2)^2}\right] = e^{-3t} \left(\frac{t \sin(2t)}{4}\right) L−1[s+3(s2+6s+13)2]=te−3tsin⁡(2t)4L^{-1}\left[\frac{s+3}{(s^2+6s+13)^2}\right] = \boxed{\frac{t e^{-3t} \sin(2t)}{4}}
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