Partial Differential Equation

According to given condition given function f(x,y) is nothing but constant function i.e. f(x,y)=3 because this is the only function whose value is 3 at any point on the boundary of unit circle and it is also satisfying Laplace equation, so
f(0,0)=3

\begin{aligned} \cos ^4 x&=\cos ^2 x\cos ^2 x\\ &=\left ( \frac{1+\cos 2x}{2} \right )\left ( \frac{1+\cos 2x}{2} \right )\\ &=\frac{1}{4}(1+2 \cos 2x+ \cos ^2 2x)\\ &=\frac{1}{4}(1+2 \cos 2x+ \frac{(1+\cos 4x)}{2})\\ &=\frac{1}{4}+\frac{\cos 2x}{2}+\frac{1}{8}+\frac{\cos 4x}{8}\\ &=\frac{3}{8}+\frac{\cos 2x}{2}+\frac{\cos 4x}{8}\\ f_4&=\frac{1}{8}=0.125\\ f_5&=0\\ \rightarrow f_4+f_5&=0.125 \end{aligned}

\begin{aligned} z&=\sin(y+it)+ \cos (y-it)\\ \frac{\partial z}{\partial y}&=\cos (y+it)-\sin (y-it)\\ \frac{\partial ^2 z}{\partial ^2 y^2}&=-\sin(y+it)- \cos (y-it)\\ \frac{\partial ^2 z}{\partial ^2 y^2}&=-z \;\;...(i)\\ \frac{\partial z}{\partial t}&=i \cos (y+it)+i\sin (y-it)\\ \frac{\partial ^2 z}{\partial ^2 t^2}&=+\sin(y+it)+ \cos (y-it)\\ \frac{\partial ^2 z}{\partial ^2 t^2}&=z\;\;...(ii)\\ &\text{Adding (i) and (ii)}\\ &\frac{\partial ^2 z}{\partial ^2 y^2}+\frac{\partial ^2 z}{\partial ^2 t^2}=0 \end{aligned}

\begin{aligned} a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\\ &=\frac{1}{\pi}\int_{-\pi}^{\pi} (x-x^2)dx\\ &=\frac{-1}{\pi}\int_{0}^{\pi}2x^2 dx\\ &=\frac{-1}{\pi}\left ( \frac{2x^3}{3} \right )_0^{\pi}\\ &=-\frac{2}{3 \pi}[\pi^3]=\frac{-2 \pi^2}{3}\\ &=-6.58 \end{aligned}

L(\sinh (at))=\frac{a}{s^2-a^2}

The Laplace's equation of continuity for two dimensional flow in a soil is expressed as:
k_x\frac{\partial^2 h}{\partial x^2}+k_z\frac{\partial^2 h}{\partial z^2}=0... for anisotropic soil [k_x\neq k_z]
:
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial z^2}=0... for isotropic soil [k_x = k_z]

\begin{aligned} L(e^{at}) &=\frac{1}{s-a} \\ L(e^{at}) &=\int_{0}^{\infty }e^{-st}e^{at}dt \\ &= \int_{0}^{\infty }e^{-(s-a)t}dt\\ &= \left.\begin{matrix} \frac{e^{-(s-a)t}}{-(s-a)} \end{matrix}\right|_{0}^{\infty }\\ &=-\frac{1}{s-a}(0-1)\\ &=\frac{1}{s-a} \end{aligned}

The function is f(x)=0
-p\lt x\leq 0
=p-x,\, 0 \lt x \lt \pi
And Fourier series is,
f\left ( x \right )=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\frac{\cos 5x}{5^{2}}+... \right ]+\left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+... \right ] ...\left ( i \right )
At x=0, (a point of discontinuity), the fourier series converges to \frac{1}{2}\left [ f\left ( 0^{-1} \right )+f\left ( 0^{+} \right ) \right ]
where f\left ( 0^{-} \right )=\lim_{x\rightarrow 0}\left ( \pi -x \right )=\pi
Hence, eq. (i), we get,
\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{1}{1^{2}}+\frac{1}{3^{2}}+... \right ]
\Rightarrow \;\; \frac{1}{1}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+...\frac{\pi ^{2}}{8}

e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}...
(By McLaurin's series expansion)

It is a standard result that
L\left ( \cosh at \right )=\frac{s}{s^{2}-a^{2}}