finding specific solution to $u_{tt}-u_{xx}=x^2-t^2$ with boundry conditions $u(x,0)=\frac{-x^4}{16}-x, \ u_t(x,0)=1, x \in R$

Question

I am analyzing the following PDE $$u_{tt}-u_{xx}=x^2-t^2$$ with boundary conditions $$u(x,0)=\frac{-x^4}{16}-x, \quad u_t(x,0)=1, \quad x \in R$$

I understand how to derive the general solution no problem, just cannot figure out how to get the specific solution which supposed to be $\ \ \mathbf{u(x,t)=\frac{-(x^2-t^2)^2}{16}+t-x}$

I got the following general solution $$u(x,t)=\frac{-(x^2-t^2)^2}{16}+F(x+t)+G(x-t)$$

applying the first condition $u(x,0)$ I get $$F(x)+G(x)=-x$$

applying second condition $u_t(x,0)=1$ I get $$F'(x)+G'(x)=1$$

how should this be continued? where the t (second term of the final specific solution) comes from?

Answer 1

$$u(x,t)=\frac{-(x^2-t^2)^2}{16}+F(x+t)+G(x-t)$$

$$u_t=\frac{-(x^2-t^2)t}{4}+F'(x+t)-G'(x-t)$$

$\begin{cases} u(x,0)=-\frac{x^4}{16}+F(x)+G(x)=-\frac{x^4}{16}-x \quad\to\quad F(x)+G(x)=-x \\ u_t(x,0)=F'(x)-G'(x)=1 \quad\to\quad F(x)-G(x)=x+c \end{cases}$ $$\begin{cases} F(x)=\frac{c}{2} \quad\to\quad F(x-t)=\frac{c}{2}\\ G(x)=-x-\frac{c}{2} \quad\to\quad G(x-t)=-(x-t)-\frac{c}{2} \end{cases}$$ $u(x,t)=\frac{-(x^2-t^2)^2}{16}+\frac{c}{2}-(x-t)-\frac{c}{2}$ $$u(x,t)=\frac{-(x^2-t^2)^2}{16}-x+t$$