Laplace Transform: (t sin(at) cos(bt))^2

How can I laplace transform this (t sin(at) cos(bt))2

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((ab)t)+12sin((a+b)t))2 =(12sin((ab)t))2+14sin((a+b)t)sin((ab)t)+14sin((a+b)t)sin((ab)t)+(12sin((a+b)t))2 =(12sin((ab)t))2+12sin((a+b)t)sin((ab)t)+(12sin((a+b)t))2 =14sin2((ab)t)+12sin((a+b)t)sin((ab)t)+14sin2((a+b)t) =14(1cos(2(ab)t)2)+12sin((a+b)t)sin((ab)t)+14(1cos(2(a+b)t)2) =18(1cos(2(ab)t))+12(12cos(((ab)(a+b))t)12cos(((ab)+(a+b))t))+18(1cos(2(a+b)t) =18(1cos(2(ab)t))+14(cos((2b)t)cos((2a)t)+18(1cos(2(a+b)t)) =1418cos(2(ab)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 1418cos(2(ab)t)+14cos((2b)t)14cos((2a)t)18(cos(2(a+b)t)) would be:

14ss8(s2+(2(ab))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 14ss8(s2+(2(ab))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(ab))2)3+6s(s2+(2(ab))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(ab))2)3+6s(s2+(2(ab))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(ab))2)3+6s(s2+(2(ab))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......