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To get the Laplace transform
To get the Laplace transform of (tsin(at)cos(bt))2, start by modifying (sin(at)cos(bt))2 to an equation where it is easier to get its Laplace transform.
(sin(at))2(cos(bt))2=(sin(at)cos(bt))2 =(12sin((a−b)t)+12sin((a+b)t))2 =(12sin((a−b)t))2+14sin((a+b)t)sin((a−b)t)+14sin((a+b)t)sin((a−b)t)+(12sin((a+b)t))2 =(12sin((a−b)t))2+12sin((a+b)t)sin((a−b)t)+(12sin((a+b)t))2 =14sin2((a−b)t)+12sin((a+b)t)sin((a−b)t)+14sin2((a+b)t) =14(1−cos(2(a−b)t)2)+12sin((a+b)t)sin((a−b)t)+14(1−cos(2(a+b)t)2) =18(1−cos(2(a−b)t))+12(12cos(((a−b)−(a+b))t)−12cos(((a−b)+(a+b))t))+18(1−cos(2(a+b)t) =18(1−cos(2(a−b)t))+14(cos((−2b)t)−cos((2a)t)+18(1−cos(2(a+b)t)) =14−18cos(2(a−b)t)+14cos((−2b)t)−14cos((2a)t)−18(cos(2(a+b)t))
The Laplace transform of cosωot u(t) is ss2+ω2o. The Laplace transform of u(t) or 1 is 1s. Then using the Linearity principle, stating that the Laplace transform of af1(t)±bf2(t) is aF1(s)±bF2(s), which is basically getting the individual Laplace transforms of individual terms on the equation.
Armed with these...the Laplace transform of 14−18cos(2(a−b)t)+14cos((−2b)t)−14cos((2a)t)−18(cos(2(a+b)t)) would be:
14s−s8(s2+(2(a−b))2)+s4(s2+(2(b))2)−s4(s2+(2(a))2)+s8(s2+(2(a+b))2)
To finally get the Laplace transform of (tsin(at)cos(bt))2, notice that the equation t2sin2(at)cos2(bt) has a factor t2. Using the property of Laplace transform involving an equation multiplied by tn, where n=1,2,3,..., the Laplace transform of equations f(t) being multiplied by tn would be (−1)ndnF(s)dsn.
In this case, the Laplace transform of t2sin2(at)cos2(bt) would be (−1)2d2F(s)ds2. We will get the expression for (−1)2d2F(s)ds2. Now getting the second derivatives of 14s−s8(s2+(2(a−b))2)+s4(s2+(2(b))2)−s4(s2+(2(a))2)+s8(s2+(2(a+b))2),
it was found out that the second derivative of the preceding equation would be:
12s3+18(8s3(s2+(2(a−b))2)3+−6s(s2+(2(a−b))2)2)+14(8s3(s2+(2(b))2)3+−6s(s2+(2(b))2)2)−14(8s3(s2+(2(a))2)3+−6s(s2+(2(a))2)2)+(8s3(s2+(2(a+b))2)3+−6s(s2+(2(a+b))2)2)
I've skipped the step where I get the second derivative. It's too long.
Therefore, the Laplace transform of t2sin2(at)cos2(bt) is (−1)2d2F(s)ds2:
L{t2sin2(at)cos2(bt)}=(−1)2(12s3+18(8s3(s2+(2(a−b))2)3+−6s(s2+(2(a−b))2)2)+14(8s3(s2+(2(b))2)3+−6s(s2+(2(b))2)2)−14(8s3(s2+(2(a))2)3+−6s(s2+(2(a))2)2)−18(8s3(s2+(2(a+b))2)3+−6s(s2+(2(a+b))2)2)) =12s3+18(8s3(s2+(2(a−b))2)3+−6s(s2+(2(a−b))2)2)+14(8s3(s2+(2(b))2)3+−6s(s2+(2(b))2)2)−14(8s3(s2+(2(a))2)3+−6s(s2+(2(a))2)2)−18(8s3(s2+(2(a+b))2)3+−6s(s2+(2(a+b))2)2).
Alternate solutions are encouraged......