91影视

It has been given to continue with problem 4.

The Laplace transform of an ordinary differential equation converts it into an algebraic equation. Taking the Laplace transform of a partial differential equation reduces the number of independent variables by one, and so converts a two-variable partial differential equation into an ordinary differential equation.

Write the expression for B(k) in the integral form.

$B\left(k\right)=\frac{200}{蟺}{鈭�}_0^1\sin \left(kz\right)dz$

Start solving.

$\begin{array}{c}u(x,t)=\frac{200}{蟺}{鈭�}_0^{\mathrm{鈭�}}B\left(k\right)e^{鈭�{伪}^2{k}^{2}t}\sin \left(kx\right)dk\\ =\frac{200}{蟺}{鈭�}_0^{\mathrm{鈭�}}{e}^{鈭�{伪}^2{k}^{2}t}\sin \left(kz\right)\sin \left(kx\right)dk{鈭�}_0^{1}dz\end{array}$

Use the following trigonometric formula.

$\sin \left(kx\right)\sin \left(kz\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.[\cos (kx鈭�kz)鈭�\cos (kx+kz)]$

It can written as mentioned below.

$u(x,t)=\frac{200}{蟺}{鈭�}_0^{\mathrm{鈭�}}{e}^{鈭�{伪}^2{k}^{2}t}/2[\cos (kx鈭�kz)鈭�\cos (kx+kz)]dk{鈭�}_0^{1}dz$

If table of integrals is checked the equation mentioned below is found.

${鈭�}_0^{\mathrm{鈭�}}{e}^{鈭�a{x}^2}\cos \left(bx\right)dx=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\sqrt{蟺/a}{e}^{鈭�b^2/\left(4a\right)}$

Thus, the equation changes.

$u(x,t)=\frac{100}{蟺}{鈭�}_0^1\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\frac{\sqrt{蟺}}{{伪}^{2}t}(e^{\frac{{(x鈭�z)}^2}{4{伪}^{2}t}}鈭�e^{\frac{{(x+z)}^2}{4{伪}^{2}t}})dz$

For the final step, check on a table of integrals again.

${鈭�}_0^{\mathrm{鈭�}}{e}^{\frac{{(a鈭�x)}^2}{b}}dx=鈭�\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\sqrt{蟺b}erf\left(\frac{a鈭�x}{\sqrt{b}}\right)$

Thus, identifying every parameter with our integral.

$u(x,t)=100\mathrm{erf}\left(\frac{x}{2伪\sqrt{t}}\right)鈭�50\mathrm{erf}\left(\frac{x鈭�1}{2伪\sqrt{t}}\right)鈭�50\mathrm{erf}\left(\frac{x+1}{2伪\sqrt{t}}\right)$

Hence, the final answer is$\mathrm{u}(\mathrm{x},\mathrm{t})=100\mathrm{erf}\left(\frac{\mathrm{x}}{2\mathrm{伪}\sqrt{\mathrm{t}}}\right)鈭�50\mathrm{erf}\left(\frac{\mathrm{x}鈭�1}{2\mathrm{伪}\sqrt{\mathrm{t}}}\right)鈭�50\mathrm{erf}\left(\frac{\mathrm{x}+1}{2\mathrm{伪}\sqrt{\mathrm{t}}}\right)$.