Inverse Laplace Transform,

Question

I have been stuck on this problem for quite a bit, have tried to look at similar answers on website but no help...

The original questions is, Solve the IVP

$\ y''+y=\sin(t);y(0)=1;y'(0)=0$

I applied Laplace to both sides and ended up with this

$\ Y(s) = \frac {s}{s^2+1} - \frac {1}{s^2+1} + \frac {1}{(s^2+1)^2}$

where $\ L(y(t)) = Y(s) $

so $\ y(t) = \cos(t) - \sin(t) +L(\frac {1}{(s^2+1)^2}) $

the first to fractions on the RHS of the equation are simple to solve, what I am having trouble with is the last fraction. I have tried partial fractions, but got nothing out of it. I am not sure where to go with this?

I tried using this solution, but wasn't sure how to apply it to my problem; Finding the inverse Laplace transform of $\frac{s^2-4s-4}{s^4+8s^2+16}$

Answer 1

The answer is $$L^{-1}\left(\frac{1}{(s^2+1)^2}\right)=\frac{\sin t-t\cos t}{2}\ .$$ You will find this in extended Laplace tables. If you want to derive it for yourself from more basic theorems you can use the following: $$L(\sin t)=\frac{1}{s^2+1}\ ,\quad L(tf(t))=-F'(s)\ ,\quad L(f'(t))=sF(s)-f(0)\ ,$$ where $F(s)$ denotes the transform of $f(t)$.

Step 1: use the first two results to find $L(t\sin t)$.

Step 2: then use the third result to find $L(t\cos t+\sin t)$.

Step 3: use the algebraic identity $$\frac{s^2}{(s^2+1)^2}=\frac{1}{s^2+1}-\frac{1}{(s^2+1)^2}$$ to finish the problem.

Answer 2

When I take the Laplace Transform of your given problem, I get $\frac{1+s+s^3}{(1+s^2)^2}$, which is not equal to the expression you got.

Using the rules

• $t f(t) \rightarrow -F'(s)$
• $\sin(\omega t) H(t)\rightarrow \frac{\omega}{s^2+\omega^2}$
• $\cos(\omega t) H(t)\rightarrow \frac{s}{s^2+\omega^2}$

you should be able to discover that $\mathcal{L}^{-1}\left(\frac{1}{(1+s^2)^2}\right) = \frac{1}{2}(\sin x - x \cos x)$ and the rest is about as hard.

Answer 3

Indeed, $$ f(t) = L^{-1}\biggl[\dfrac{1}{(s^2 + 1)^2}\biggr] = L^{-1}\biggl[\dfrac{1}{s^2 + 1} - \dfrac{s^2}{(s^2 + 1)^2}\biggr] = \sin t - L^{-1}\biggl[\dfrac{s^2}{(s^2 + 1)^2}\biggr] \ (1) $$ But, $$ \dfrac{d}{ds}\biggl(\dfrac{s}{s^2 +1}\biggr) = \dfrac{1}{(s^2 + 1)^2} - \dfrac{s^2}{(s^2 + 1)^2} \quad \Rightarrow $$ $$ L^{-1}\biggl[\dfrac{d}{ds}\biggl(\dfrac{s}{s^2 +1}\biggr) \biggr] = L^{-1}\biggl[\dfrac{1}{(s^2 + 1)^2}\biggr] - L^{-1}\biggl[\dfrac{s^2}{(s^2 + 1)^2}\biggr] \quad \Rightarrow $$ $$ -t\cos t - f(t) = - L^{-1}\biggl[\dfrac{s^2}{(s^2 + 1)^2}\biggr] \quad (2) $$ Replacing (2) in the expression (1), we have $$ f(t) = \sin t -t\cos t - f(t) \quad \Rightarrow \quad f(t) = \dfrac{\sin t - t\cos t}{2} $$ Another way is to use convolution: $$ L^{-1}\biggl[\frac{1}{(1 + s^2)^2}\biggr] = \sin t \ast \sin t = \int_{0}^{t}\sin u\sin(t - u)du $$