I'm trying to find the general solution of $$\frac{\partial^2 u}{\partial xt}=6x+12t^{2}$$, where $u=u(x,t)$, subject to $u(x,2)=5$ and $u(1,t)=2t+1$.
My solution is as follows;
First integrating wrt $t$ gives $$\frac{\partial u}{\partial x}=\int(6x+12t^{2})dt=6xt+4t^{3}+f(x)$$ thus $$u(x,t)=3x^{2}t+4t^{3}x+f_{1}(x)+g(t)$$ Since we integrate $\frac{\partial u}{\partial x}$ with respect to $x$ this time, and $\int f(x)dx=f_{1}(x)+g(t)$ where $f(x),f_{1}(x),g(t)$ are arbitrary functions. I'm unsure how to apply the initial conditions from this step. I assumed that for the first condition I would just let $u(x,t)=5$ and substitute the value of $2$ in for t$ but then I still have those functions in the solution. I was wondering if anyone would be able to help clarify what I need to do and how to continue.
$$u(x,t)=3x^{2}t+4t^{3}x+f(x)+g(t)\quad\text{is OK.}$$ First condition, $u(x,2)=5$ $$6x^{2}+32x+f(x)+g(2)=5$$ $$f(x)= -6x^{2}-32x+5-g(2)$$ $$u(x,t)=3x^{2}t+4t^{3}x-6x^{2}-32x+5-g(2)+g(t)$$ Second condition, $u(1,t)=2t+1$ $$3t+4t^{3}-6-32+5-g(2)+g(t)=2t+1$$ $$g(t)=2t+1-3t-4t^3+6+32-5+g(2)$$ $$g(t)= -4t^3-t+34+g(2)$$ $$u(x,t)=3x^{2}t+4t^{3}x-6x^{2}-32x+5-g(2)-4t^3-t+34$$ $$u(x,t)=3x^{2}t+4t^{3}x-6x^{2}-32x+5-g(2)-4t^3-t+34+g(2)$$ Solution : $$u(x,t)=3x^{2}t+4t^{3}x-6x^{2}-32x-4t^3-t+39$$