91Ó°ÊÓ

The given equation is mentioned below.

$u(x,y)=\frac{200}{\mathrm{π}}{∫}_0^{\mathrm{∞}}{e}^{−ky}\sin \left(kx\right)dk{∫}_0^1\sin \left(kt\right)dt$

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 integral form of B(k).

$B\left(k\right)=\frac{2}{\mathrm{π}}{∫}_0^{1}100\sin \left(kt\right)dt$ ….. (1)

Thus, the equation obtained.

$u(x,y)=\frac{200}{\mathrm{π}}{∫}_0^{\mathrm{∞}}{∫}_0^1{e}^{−ky}\sin \left(kx\right)\sin \left(kt\right)dkdt$ ….. (2)

Remember that the variable of the integral is silent, thus it can be changed as out requirement. Also, integrate the variables in the order that is required. So, integrate first with respect to k. But, before it is required to arrange the expression by writing the product of sines as a difference of cosines (which is a trigonometric identity).

$\sin \mathrm{Î\pm }\sin \mathrm{β}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.(\cos \mathrm{Î\pm }−\mathrm{β}−\cos \mathrm{Î\pm }+\mathrm{β})$..... (3)

$u(x,y)={∫}_0^{\mathrm{∞}}{e}^{−ky}[\cos (kx−kt)−\cos (kx+kt)]dk{∫}_0^{1}dt$ ….. (4)

This integral has the form mentioned below.

${∫}_0^{\mathrm{∞}}{e}^{ax}\cos \left(bx\right)dx=\frac{a}{a^2{b}^2}$ ….. (5)

Thus, it can be written as mentioned below.

$u(x,y)=\raisebox{1ex}{$100$}\!\left/ \!\raisebox{-1ex}{$\mathrm{π}$}\right.{∫}_0^1(\frac{−y}{y^2+{(x−t)}^2}−\frac{−y}{y^2+{(x+t)}^2})dt$ ….. (6)

By checking a table of integrals, see that this integral can be splitted by two integrals with the following form.

${∫}_0^1\frac{a}{a^2+{(x+b)}^2}dx=\arctan (b+x/a)$ ….. (7)

Use eq. (7), solve eq. (6)

$u(x,y)=\frac{100}{\mathrm{π}}{[\arctan (x−t/y)+\arctan (x+t/y)]}_0^1$ ….. (8)

Evaluate the final solution the desired expression is derived.

$u(x,y)=\frac{100}{\mathrm{π}}[\arctan \left(\frac{x}{y}\right)−\arctan \left(\frac{x+1}{y}\right)−\arctan \left(\frac{x−1}{y}\right)]$