Let $u(x,t)$ be a function that satisfies $$u_{xx}-u_{tt}=e^{x}+6t, \ x \in \mathbb{R},\ t>0$$ and the initial conditions $$u(x,0)= sinx,\ \ u_{t}(x,0)=0, \ \forall \ x \in \mathbb{R}.$$
Then the value of $u(\frac{π}{2},\frac{π}{2})$ is1)$e^{\frac{π}{2}}\big(1+\dfrac{e^\frac{{π}}{2}}{2}\big)+\big(\dfrac{π^{3}+4}{8}\big)$
2)$e^{\frac{π}{2}}\big(1+\dfrac{e^\frac{{π}}{2}}{2}\big)+\big(\dfrac{π^{3}-4}{8}\big)$
3)$e^{\frac{π}{2}}\big(1-\dfrac{e^\frac{{π}}{2}}{2}\big)-\big(\dfrac{π^{3}+4}{8}\big)$
4)$e^{\frac{π}{2}}\big(1-\dfrac{e^\frac{{π}}{2}}{2}\big)-\big(\dfrac{π^{3}-4}{8}\big)$
Solution to the homogeneous pde $u_{xx}-u_{yy}=0$ is given by d'Alembert's solution to wave equation which gives $$u(x,t)= \sin x \cos t.$$
Now the particular integral of the given pde is
$\dfrac{1}{D^{2}-D'^{2}} e^{x}+6t$
=$\big(\dfrac{1}{D^{2}-D'^{2}} e^{x}\big)+
\big(\dfrac{1}{D^{2}-D'^{2}}6t\big)$
=$e^{x}+\dfrac{1}{-D'^{2}\big(1-\big(\frac{D}{D'}\big)^{2}\big)}6t$
=$e^{x}-t^{3}$
Therefore $u(x,t)=\sin x \cos t+e^{x}-t^{3}$ which does not satisfy any of the given options. Can anyone help me see where I am wrong?
Best Answer
Transforming Laplace
$$ U_{xx}(x,s)-s^2U(x,s)+\sin x = \frac 1s e^x+\frac{6}{s^2} $$
now solving for $x$
$$ U(x,s) = c_1(s) e^{s x}+c_2(s) e^{-s x}+\frac{\left(s^2-1\right) s^4 \sin (x)-\left(s^2+1\right) \left(s^2 \left(s e^x+6\right)-6\right)}{s^4 \left(s^4-1\right)} $$
Now considering $c_1(s)=c_2(s) = 0$ and antitransforming we have
$$ u(x,t) = \sin (t) \sin (x)-\cosh (t) \cosh (x)-\cosh (t) \sinh (x)+\sinh (x)+\cosh (x)-t^3 $$
hence
$$ u\left(\frac{\pi}{2},\frac{\pi}{2}\right) = \frac{1}{8} \left(4+8 e^{\pi /2}-4 e^{\pi }-\pi ^3\right) $$