Question

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]

${\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t}$

Short answer

$\mathcal{L}\left\{t^2+e^t\sin 2t\right\}=\frac{2}{s^3}+\frac{2}{{(s-1)}^2+4}$

Step-by-step solution

Find the Laplace transform of the given function

Consider the function $t^2+e^t\sin 2t$

Since $\mathcal{L}\left\{a{f}_1+b{f}_2\right\}=a\mathcal{L}\left\{f_1\right\}+b\mathcal{L}\left\{f_2\right\}$, $\mathcal{L}\left\{t^2+e^t\sin 2t\right\}=\mathcal{L}\left\{t^2\right\}+\mathcal{L}\left\{e^t\sin 2t\right\}$.

Since $\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}}$and $\mathcal{L}\left\{e^{at}\sin bt\right\}=\frac{b}{{(s-a)}^2+b^2}$, $\mathcal{L}\left\{t^2\right\}+\mathcal{L}\left\{e^t\sin 2t\right\}=\frac{2!}{s^{2+1}}+\frac{2}{{(s-1)}^2+2^2}$

Therefore, $\mathcal{L}\left\{t^2+e^t\sin 2t\right\}=\frac{2!}{s^{2+1}}+\frac{2}{{(s-1)}^2+2^2}$because $\mathcal{L}\left\{t^2+e^t\sin 2t\right\}=\mathcal{L}\left\{t^2\right\}+\mathcal{L}\left\{e^t\sin 2t\right\}$.

Hence,$\mathcal{L}\left\{t^2+e^t\sin 2t\right\}=\frac{2}{s^3}+\frac{2}{{(s-1)}^2+4}$.