Finding Laplace transform of two functions multiplied together

Question

How can I find the Laplace of

$f(t) = \cos^2(t)$

$f(t) = \sin3t\cos3t$

$f(t) = te^{t}$

$f(t) = t\cos(2t)$

What about inverse transform for

$F(s) = \frac{5-3s}{2^s+9}$

$F(s) = \frac{10s-3}{25-s^2}$

$F(s) = 2s^{-1}e^{-3s}$

Is there a general method used when you're multiplying two functions together, or have what appears to be a combination in the inverse Laplace? I was hoping I could look them up on a table of transforms, but I'm not exactly sure how to deal with them.

Answer 1

As $L(e^{at})=\frac1{s-a}$

So putting $a=0,L(1)=\frac1s$

and putting $a=c+id,L(e^{(c+id)t})=\frac1{s-(c+id)}$

so, $L(e^{ct}\cos dt)+iL(e^{ct}\sin dt)=\frac{s-c+id}{(s-c)^2+d^2}$

$cos^2t=\frac{1+\cos2t}2$

So, $L(cos^2t)=\frac12 L(1)+\frac12 L(\cos2t)=\frac1{2s}+\frac s{2(s^2+2^2)}$

$\cos3t\sin3t=\frac{\sin6t}2$

So, $L(\cos3t\sin3t)=\frac{L(\sin6t)}2=\frac{6}{2(s^2+6^2)}$

We know, $L(e^{at}f(t))=F(s-a)$ where $F(s)=L(f(t))$

So, $L(e^{at} t^n)=\frac{n!}{(s-a)^{n+1}}$

Hence $L(te^t)=\frac1{(s-a)^2}$ putting $a=n=1$

Putting $a=2i,n=1; L(e^{2it} t)=\frac1{(s-2i)^2}$

$L(t\cos 2t)+iL(t\sin 2t)=\frac{(s+2i)^2}{(s^2+4)^2}=\frac{s^2-4+i4s}{(s^2+4)^2}$

Now compare the real and the imaginary parts.

Answer 2

Use trig identities to help out with integration: $(cos(t))^2 = (1/2)(cos(2t) + 1)$

Which leads to: $L{(cos(t))^2} = L{(1/2)(cos(2t) + 1)} = (1/2)L{(cos(2t))} + (1/2)L{(1)}$

But we now have, $L{(1)} = L{(H(t))} = 1/s$ and $ L{(cos(2t))} = (1/2)L{(cos(t))}(s/2) = (1/2)(\frac{s}{4((s^2)} + 1) = 2s/(s^2 + 4)$

Finally:

$$L{(cos(t))^2} = s/(s^2 + 4) + (1/2s) = (3s^2 + 4)/(2s)(s^4 + 4)$$