0
$\begingroup$

I am a math student at freshman university level and I am taking an ODE class. I am not expected to integrate in the complex plane or to know convolution (so no duplication with Inverse Laplace of $\frac 1 {(s^2+a^2)^n}$). The way we've done inverse Laplace in this class is by using the basic properties of Laplace Transform and referring to known Laplace transforms like sin, cos, and polynomials)

I don't know where to start but I know a first step is to consider that $$\mathcal{L} (t\times \cos t)= -\frac{d}{ds}\frac{s}{s^2+1}$$

$\endgroup$

1 Answer 1

1
$\begingroup$

Hint

You have that $$\mathcal{L} (t\times \cos t)= -\frac{d}{ds}\frac{s}{s^2+1}=\frac {s^2-1}{(s^2+1)^2}$$ $$\mathcal{L} (\sin t)= \frac{1}{s^2+1}=\frac {s^2+1}{(s^2+1)^2}$$

Substract ...

$\endgroup$
2
  • $\begingroup$ oh I see, If i subtract, I get $$ \frac{-2}{s^2+1)^2 $$ and can just take the constant out $\endgroup$
    – JohnA.
    Jun 6, 2018 at 5:14
  • 1
    $\begingroup$ its just a constant....Laplace transform is a linear operator @JohnA. $L(af)=aL(f)$ $\endgroup$ Jun 6, 2018 at 5:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .