Find L⁻¹\[\(1)/(s(s+1)(s+2))\]

Here are the solutions for questions 7, 10, 11, and 12:

7. Find $L^{-1}\left[\frac{1}{s(s+1)(s+2)}\right]$

Step 1: Perform partial fraction decomposition.
Let $\frac{1}{s(s+1)(s+2)} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+2}$.
Multiply both sides by $s(s+1)(s+2)$:

$$1 = A(s+1)(s+2) + B s(s+2) + C s(s+1)$$

Step 2: Solve for the constants $A$, $B$, and $C$.

• Set $s=0$: $1 = A(1)(2) \implies 1 = 2A \implies A = \frac{1}{2}$
• Set $s=-1$: $1 = B(-1)(-1+2) \implies 1 = B(-1)(1) \implies B = -1$
• Set $s=-2$: $1 = C(-2)(-2+1) \implies 1 = C(-2)(-1) \implies 1 = 2C \implies C = \frac{1}{2}$

Step 3: Substitute the constants back into the partial fraction expansion.

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

Step 4: Apply the inverse Laplace transform to each term.
Using the formulas $L^{-1}\left[\frac{1}{s}\right] = 1$ and $L^{-1}\left[\frac{1}{s-k}\right] = e^{kt}$:

$$L^{-1}\left[\frac{1}{s(s+1)(s+2)}\right] = \frac{1}{2}L^{-1}\left[\frac{1}{s}\right] - L^{-1}\left[\frac{1}{s+1}\right] + \frac{1}{2}L^{-1}\left[\frac{1}{s+2}\right]$$ $$L^{-1}\left[\frac{1}{s(s+1)(s+2)}\right] = \frac{1}{2}(1) - e^{-t} + \frac{1}{2}e^{-2t}$$ $$L^{-1}\left[\frac{1}{s(s+1)(s+2)}\right] = \boxed{\frac{1}{2} - e^{-t} + \frac{1}{2}e^{-2t}}$$

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10. Find $L[t^3 e^{at}]$

Step 1: Identify the function $f(t)$ and the power $n$.
Here, $f(t) = e^{at}$ and $n=3$.

Step 2: Find the Laplace transform of $f(t)$.

$$F(s) = L[e^{at}] = \frac{1}{s-a}$$

Step 3: Apply the property $L[t^n f(t)] = (-1)^n \frac{d^n}{ds^n} F(s)$.
For $n=3$:

$$L[t^3 e^{at}] = (-1)^3 \frac{d^3}{ds^3}\left(\frac{1}{s-a}\right)$$ $$L[t^3 e^{at}] = - \frac{d^3}{ds^3}(s-a)^{-1}$$

Step 4: Calculate the derivatives.
First derivative: $\frac{d}{ds}(s-a)^{-1} = -1(s-a)^{-2}$
Second derivative: $\frac{d^2}{ds^2}(s-a)^{-1} = (-1)(-2)(s-a)^{-3} = 2(s-a)^{-3}$
Third derivative: $\frac{d^3}{ds^3}(s-a)^{-1} = 2(-3)(s-a)^{-4} = -6(s-a)^{-4}$

Step 5: Substitute the third derivative back into the formula.

$$L[t^3 e^{at}] = - (-6(s-a)^{-4})$$ $$L[t^3 e^{at}] = 6(s-a)^{-4}$$ $$L[t^3 e^{at}] = \boxed{\frac{6}{(s-a)^4}}$$

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11. Find $L[t e^t \sin t]$

Step 1: Find the Laplace transform of $\sin t$.
Let $f(t) = \sin t$.

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

Step 2: Apply the First Shifting Theorem for $L[e^t \sin t]$.
If $L[f(t)] = F(s)$, then $L[e^{at}f(t)] = F(s-a)$. Here $a=1$.

$$L[e^t \sin t] = \frac{1}{(s-1)^2+1}$$

Let $G(s) = \frac{1}{(s-1)^2+1}$.

Step 3: Apply the multiplication by $t$ property for $L[t (e^t \sin t)]$.
If $L[g(t)] = G(s)$, then $L[t g(t)] = - \frac{d}{ds} G(s)$.

$$L[t e^t \sin t] = - \frac{d}{ds}\left(\frac{1}{(s-1)^2+1}\right)$$ $$L[t e^t \sin t] = - \frac{d}{ds}((s-1)^2+1)^{-1}$$

Step 4: Calculate the derivative.

$$\frac{d}{ds}((s-1)^2+1)^{-1} = -1((s-1)^2+1)^{-2} \cdot \frac{d}{ds}((s-1)^2+1)$$ $$= -1((s-1)^2+1)^{-2} \cdot (2(s-1))$$ $$= -\frac{2(s-1)}{((s-1)^2+1)^2}$$

Step 5: Substitute the derivative back into the formula.

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