Second-order differential equation with initial conditions

Question

I try to solve using laplace transform $y'' + y = \sin (t)$ with $y(0) = 1$ and $y'(0) = 2$ but I don't get a solution, I don't know why. I check my work and it seems fine.

My calculations

Is possible the initial conditions are wrong, or not consistent with the system.

Answer 1

After the transform we have

$$ s^2Y(s)+Y(s) = \frac{1}{s^2+1}+s \dot y(0) + y(0) $$

or

$$ Y(s) = \frac{1}{(s^2+1)^2}+\frac{s}{s^2+1}\dot y(0)+\frac{1}{s^2+1}y(0) $$

At this point we can use the Laplace transforms table assigning the due inverses

$$ y(t) = \left(\frac{1}{2}(\sin(t)-t\cos(t))+\cos(t)\dot y(0)+\sin(t)y(0)\right)u(t) $$

where $u(t)$ is the unit step function.

Answer 2

You made some mistakes for B and C

$$B=\frac 1 {s+1} \quad ;C=\frac 3 {s(s+1]}$$

You already had some mistakes before that step... It should be

$$F(s)=\frac s {(s^2+1)}+\frac 1 {(s^2+1)^2}+\frac 2 {s^2+1}$$ You applied the laplace transform on $$y''+y'=\sin(t)$$ And not on $$y''+y=\sin(t)$$