Question

In Problems 43-46 use Problems 41 and 42 and the fact that$\mathit{\Gamma }\left(\frac{1}{2}\right)\mathbf{=}\sqrt{\mathbf{\pi }}$ to find the Laplace transform of the given function.

46.$f\left(t\right)=2t^{1/2}+8t^{5/2}$

Short answer

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{\sqrt{\pi }}{s^{\frac{3}{2}}}+\frac{15\sqrt{\pi }}{s^{\frac{7}{2}}}$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $t⩾0$. Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of f, provided that the integral converges.

 Use Laplace Transform

Take Laplace transform of both side.

$\mathcal{L}\left\{f\right(t\left)\right\}=\mathcal{L}\left\{2t^{\frac{1}{2}}+8t^{\frac{5}{2}}\right\}(\mathcal{L}\{\alpha f\left(t\right)+\beta g\left(t\right)\}=\alpha \mathcal{L}\{f\left(t\right)\}+\beta \mathcal{L}\{g\left(t\right)\left\}\right)$$$

$=2\mathcal{L}\left\{t^{\frac{1}{2}}\right\}+8\mathcal{L}\left\{t^{\frac{5}{2}}\right\}$----------- (1)

First we find$\mathcal{L}\left\{t^{\frac{1}{2}}\right\}$.

$\mathcal{L}\left\{t^{\frac{1}{2}}\right\}=\frac{\Gamma \left(\frac{1}{2}+1\right)}{s^{\frac{1}{2}}1}\left(\mathcal{L}\left\{t^{\alpha }\right\}=\frac{\Gamma (\alpha +1)}{s^{\alpha +1}}\right)$

$=\frac{\frac{1}{2}\Gamma \left(\frac{1}{2}\right)}{s^{\frac{3}{2}}}\left(\Gamma \right(\alpha +1)=\alpha \Gamma (\alpha \left)\right)$

$=\frac{\sqrt{\pi }}{2s^{\frac{3}{2}}}\left(\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }\right)$

Therefore,$\mathcal{L}\left\{t^{\frac{1}{2}}\right\}=\frac{\sqrt{\pi }}{2s^{\frac{3}{2}}}$----------- (2)

Next we find$\mathcal{L}\left\{t^{\frac{5}{2}}\right\}$.

$\mathcal{L}\left\{t^{\frac{5}{2}}\right\}=\frac{\Gamma \left(\frac{5}{2}+1\right)}{s^{\frac{5}{2}+1}}\left(\mathcal{L}\left\{t^{\alpha }\right\}=\frac{\Gamma (\alpha +1)}{s^{\alpha +1}}\right)$

$=\frac{\frac{5}{2}\Gamma \left(\frac{5}{2}\right)}{s^{\frac{7}{2}}}\left(\Gamma \right(\alpha +1)=\alpha \Gamma (\alpha \left)\right)$

$=\frac{\frac{5}{2}\Gamma \left(\frac{3}{2}+1\right)}{s^{\frac{7}{2}}}\left(\Gamma \right(\alpha +1)=\alpha \Gamma (\alpha \left)\right)$

$=\frac{\frac{5}{2}\cdot \frac{3}{2}\Gamma \left(\frac{3}{2}\right)}{s^{\frac{7}{2}}}$

$=\frac{\frac{5}{2}·\frac{3}{2}\Gamma \left(\frac{1}{2}+1\right)}{s^{\frac{7}{2}}}\left(\Gamma \right(\alpha +1)=\alpha \Gamma (\alpha \left)\right)$

$=\frac{15\sqrt{\pi }}{8s^{\frac{7}{2}}}\left(\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi }\right)$

Therefore,$\mathcal{L}\left\{t^{\frac{5}{2}}\right\}=\frac{15\sqrt{\pi }}{8s^{\frac{7}{2}}}$ ----------- (3)

Substitute (2) and (3) in (1), to get

$\mathcal{L}\left\{2t^{\frac{1}{2}}+8t^{\frac{5}{2}}\right\}=2\mathcal{L}\left\{t^{\frac{1}{2}}\right\}+8\mathcal{L}\left\{t^{\frac{5}{2}}\right\}$

$=\frac{2\sqrt{\pi }}{2s^{\frac{3}{2}}}+\frac{\$·15\sqrt{\pi }}{\$s^{\frac{7}{2}}}$

$=\frac{\sqrt{\pi }}{s^{\frac{3}{2}}}+\frac{15\sqrt{\pi }}{s^{\frac{7}{2}}}$

Therefore,$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{\sqrt{\pi }}{s^{\frac{3}{2}}}+\frac{15\sqrt{\pi }}{s^{\frac{7}{2}}}$