Question

A) laplace transform of 4t^2 sin 3t + e^-2t + t
b)......................... te^-2t cos 3t

Answer 1

a.L(s)=
`(-24)(81-s^4)/((s^2+9)^4)+(1)/(s+2)+(1)/(s+1)`

b.
`L(s)=-(-s^2-4s+5)/((s^2+4s+13)^2)`

Explanation:

We have to find the laplace transform of each function

a.
`4t^2sin3t+e^(-2t)+t`

`L(t^nf(t))=(-1)^n(d^nF(s))/(ds^n)`

`L(t^n)=(n!)/(s^n+1)`

`L(e^(at))=(1)/(s-a)`

`L(sinat)=(a)/(s^2+a^2)`

`L(e^(at)cosbt)=(s-a)/((s-a)^2+b^2)`

Apply the formula

Then we get

`12(-1)^2(d^2((1)/(s^2+9)))/(ds^2)+(1)/(s+2)+(1)/(s+1)`

`(-24)(s^4+81+18s^2-2s^4-18s^2)/((s^2+9)^4)+(1)/(s+2)+(1)/(s+1)`

L(s)=
`(-24)(81-s^4)/((s^2+9)^4)+(1)/(s+2)+(1)/(s+1)`

b.
`te^(-2t)cos3t`

f(t)=
`e^(-2t)cos3t`

`F(s)=(s+2)/((s+2)^2+9)=(s+2)/(s^2+4s+13)`

`L(s)=-(dF(s))/(ds)=-(s^2+4s+13-(2s+4)(s+2))/((s^2+4s+13)^2)`

`L(s)=-(5-s^2-4s)/((s^2+4s+13)^2)`