To find the Laplace transform of the differential equation

$y'' + 2y' + y = \sin t$

we proceed as follows:

Step 1: Take the Laplace Transform of Each Term

Using the linearity property of the Laplace transform, we apply the transform to each term separately.

Transform of $y''$:

$\mathcal{L}\{y''\} = s^2 Y(s) - s y(0) - y'(0)$

Transform of $2y'$:

$\mathcal{L}\{2y'\} = 2(s Y(s) - y(0))$

Transform of $y$:

$\mathcal{L}\{y\} = Y(s)$

Transform of $\sin t$:

$\mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1}$

Step 2: Substitute These Transforms into the Equation

Now substitute each transform into the equation:

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

Step 3: Group Terms Involving $Y(s)$

Combine all terms involving $Y(s)$:

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

The expression $s^2 + 2s + 1$ simplifies to $(s + 1)^2$, so we have:

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

Step 4: Solve for $Y(s)$

Assuming initial conditions $y(0) = 0$ and $y'(0) = 0$, the equation simplifies further:

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

Therefore:

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

This is the Laplace transform $Y(s)$ of the solution $y(t)$ for the differential equation.

Would you like me to proceed with partial fraction decomposition to find the inverse Laplace transform and obtain $y(t)$?

Here are some related questions:

1. How does the Laplace transform simplify solving differential equations?
2. What is the significance of initial conditions in Laplace transforms?
3. How would the process change if the forcing function was $\cos t$ instead?
4. What happens if we have non-zero initial conditions?
5. Why is partial fraction decomposition useful for inverse Laplace transforms?

Tip: Always confirm initial conditions before applying the Laplace transform in differential equations; they affect the solution form